EARTH  SLOPES,REL\INING 
WALLS  AND  DAMS 


CHARLES  PRELINI 


D.VAN  NOSTRAND  COMPANY 
NEW  YORK 


LIBRARY 

OF  THE 

UNIVERSITY  OF  CALIFORNIA. 

Class 


GRAPHICAL  DETERMINATION  OF  EARTH 
SLOPES,  RETAINING   WALLS   AND   DAMS 


GRAPHICAL   DETERMINATION 

OF 

EARTH  SLOPES,  RETAINING 
WALLS  AND   DAMS 


BY 

CHARLES  PRELINI,  C.E. 

/  M 

PROFESSOR  OF  CIVIL  ENGINEERING,  MANHATTAN  COLLEGE,  NEW  YORK  CITY 

Author  of  "Earth  and  Rock  Excavation,"  "  Tunneling,"  etc. 


NEW  YORK 

D.   VAN    NOSTRAND    COMPANY 

23  MURRAY  AND  1908  27  WARREN  STS 


Copyright,  1908 
BY  D.  VAN  NOSTRAND  COMPANY 


BEHERJH, 


The  Plimpton  Press  Norwood  Mass.  U.S.A. 


PREFACE 

A  large  part  of  this  work  consists  of  graphical  methods  of 
solving  problems  concerning  the  slopes  of  earth  embankments, 
the  lateral  pressure  of  earth  against  a  wall,  and  the  thickness 
of  retaining  walls  and  dams.  The  graphical  methods  of 
Culmann,  Rebhann,  Weyrauch,  Blanc,  and  others  have  been 
employed ;  the  general  course  of  the  discussion  is  similar 
to  that  followed  by  Professor  Senesi,  of  Italy.  Hence,  with 
the  exception  of  the  graphical  determination  of  earth  slopes 
of  uniform  stability,  there  is  nothing  in  these  pages  which 
has  not  been  already  published  and  criticised.  But  the 
prominence  given  to  the  graphical  over  the  analytical  mode 
of  treatment  may  be  found  useful  to  a  numerous  class  of 
students. 

The  book  is  divided  into  five  chapters.  In  the  first  chap- 
ter attention  is  given  to  the  forces  which  determine  the 
various  slopes  of  the  earth  embankments.  Young  engineers 
will  find  it  profitable  to  study  with  care  the  economies 
secured  by  designing  the  slopes  of  deep  trenches  for  equal 
stability  instead  of  using  the  ordinary  slope  of  1  to  1. 

The  second  chapter  is  devoted  to  the  graphical  determi- 
nation of  the  pressure  of  earth  against  a  retaining  wall,  fol- 
lowing the  theory  of  Professor  Rebhann.  This  chapter  also 
contains  solutions  of  the  different  problems  which  may  be 
encountered  in  practical  work. 

In  the  third  chapter  is  given  the  analytical  demonstration 

of  Professor  Rebhann's  theory,  together  with  the  formulas 

iii 


176746 


iv  PREFACE 

deduced  from  the  analytical  theories  of  Weyrauch  and  Ran- 
kine.  In  three  tables  a  comparison  is  made  between  the 
results  obtained  by  the  graphical  process  of  determining  the 
earth  pressure  and  those  given  by  the  analytical  method. 
As  will  be  seen,  the  results  agree  well,  showing  that  for 
practical  purposes  both  methods  lead  to  almost  the  same 
result.  The  tables  were  deduced  from  the  class  work  of 
the  author's  students  at  Manhattan  College. 

The  fourth  chapter  is  devoted  to  the  design  of  retaining 
walls.  Here  will  be  found  the  most  common  types  of  wall 
used  in  practice,  together  with  the  manner  of  determining 
the  thickness  of  their  bases  both  graphically  and  analytically. 

The  fifth  chapter  is  devoted  to  dams.  The  space  devoted 
to  a  subject  so  popular  and  extensively  discussed  in  schools 
and  text-books  may  seem  disproportionately  small ;  but  in 
the  class  room,  dams  should  be  taken  for  what  they  are; 
viz.  a  particular  case  of  retaining  walls  in  which  the  mate- 
rial to  be  sustained  is  deprived  of  friction. 

The  discussion  of  the  reliability  of  the  various  theories 
is  omitted  in  order  to  avoid  confusion  in  the  untrained  mind 
of  the  student,  who  is  unable  to  follow  with  profit  such  com- 
plicated discussions  based  upon  slight  differences  in  -the 
assumptions.  It  is  only  when  students  have  mastered  the 
subject  that  they  are  able  to  discuss  intelligently  the  various 
theories  and  to  accept  or  discard  them  according  to  their 
comparative  value.  Such  critical  work  should  be  done  by 
the  students  individually  and  not  collectively.  To  show 
young  men  at  the  very  beginning  of  their  professional  studies 
that  all  theories  are  more  or  less  defective  would  generate 
confusion,  produce  uncertainty  and  a  want  of  self-confidence. 
It  would  lead  to  a  depreciation  of  every  theory  and  to  undue 
reliance  on  practical  formulas. 


PREFACE  V 

In  conclusion,  these  pages  are  intended  for  students  and 
not  for  professional  engineers.  Simplicity  and  clearness 
have  been  the  main  objects  in  view;  the  experience  of  the 
class  room  makes  the  author  believe  that  this  little  work 
will  be  of  use  to  students  and  teachers,  while  at  the  same 
time  it  may  be  of  some  help  to  the  practical  engineer. 

C.  PRELINI. 

MANHATTAN  COLLEGE,  NEW  YORK, 

September,  1908. 


CONTENTS 

CHAPTER  !• 
THE  STABILITY  OF  EARTH  SLOPES 

SECTION  PAGE 

1.  Measurement  of  Slopes 1 

2.  Equilibrium  of  a  Slope 2 

3.  Weight  of  Soils ;  Specific  Weight       .        .        .                 .        .  2 

4.  Internal  Friction  of  Soils ;  Natural  Slope,  or  Slope  of  Repose  .  3 

5.  Cohesion  in  Soil ;  The  Coefficient  of  Cohesion   ....  5 

6.  Graphical  Determination  of  the  Coefficient  of  Cohesion     .         .  7 

7.  The    Parabola    of    Cohesion ;    Practical   Applications.      The 

Stepped  Slope.     The  Slope  of  Equal  Stability        ...  12 

8.  Analytical  Calculation  of  Cohesion ;  Slope  without  Surcharge  .  18 

9.  Surcharged  Slope 20 

10.  Earth  Slopes  in  Practice 22 

CHAPTER  II 
RETAINING  WALLS  :   GRAPHICAL  METHODS 

11.  Theories  of  Earth  Pressure.     Theory  of  the  Sliding  Prism,  or 

Coulomb's  Theory.     Analytical  Theory,  or  Rankine's  Theory  28 

12.  Graphical  Method  after  Rebhann .30 

13.  Values  of  <£  and  <£' 33 

14.  Location  of  Plane  of  Rupture      .         .         .         .  -       .         .        .33 

15.  The  Triangle  of  Pressure 37 

16.  Application  of  the  Method  to  Various  Practical  Cases       .        .  38 

17.  Surcharged  Embankment;  Any  Angle  of  Surcharge          .        .  39 

18.  Case  of  No  Surcharge 44 

19.  Embankment  with  Maximum  Surcharge ;  Angle  of  Surcharge 

Equal  to  Angle  of  Repose .46 

20.  Embankment  with  Irregular  Surcharge ;  Top  of  Embankment 

a  Polygonal  Profile 46 

21.  Embankment  with  Irregular  Surcharge;  Top  of  Embankment 

of  Curvilinear  Profile 48 

vii 


viii  CONTENTS 


22.  Variation  of  Pressure  with   Height   of  Wall ;     Intensity  of 

Pressure  ;  Center  of  Pressure  .......  50 

23.  Intensity  of  Pressure    .........  52 

24.  Center  of  Pressure        .........  55 

25.  The  Earth  Pressure  represented  by  a  Line          »        .         .         .55 

26.  Effect  of  Cohesion  on  Pressure  against  Retaining  Walls    .         .  58 

27.  The  Pressure  of  Passive  Resistance  of  the  Earth  62 


CHAPTER  III 

RETAINING  WALLS  (Continued}.     ANALYTICAL  METHODS 

28.  Rebhann's  Analytical  Method 67 

29.  Formulas  of  Rankine  and  Weyrauch  ......       74 

30.  Comparison  of  Graphical  Method  with  Formulas  of  Rankine 

and  Weyrauch .        .  _  . ,- •»   :     . 78 


CHAPTER   IV 

THE  DESIGN  OF  RETAINING  WALLS 

31.  Types  of  Retaining  Walls 82 

32.  Plain  Retaining  Walls         . 82 

33.  Retaining  Walls  with  Counterforts 84 

34.  Retaining  Walls  with  Buttresses       . 84 

35.  The  Equilibrium  of  Retaining  Walls  ......  85 

36.  Determination  of  Width  of  Base  by  Graphical  Methods    .         .  88 

37.  Retaining  Wall  with  Vertical  Front  and  Back  ....  89 

38.  Retaining  Wall  with  Vertical  Front  and  Inclined  Back     .         .  91 

39.  Retaining  Wall  with  Inclined  Front  and  Vertical  Back    .         .  91 

40.  Retaining  Wall  with  Faces  inclined  in  Opposite  Directions       .  92 

41.  Retaining  Wall  with  Parallel  Inclined  Faces      ....  93 

42.  Interpolation  Method .         .93 

43.  Retaining  Walls  with  Counterforts     .         .        ...         .95 

a.   Thickness  of  Wall  Given    .         .         ...         .         .95 

1).   Thickness  of  Counterfort  Given          .        .        .        .         .97 

44.  Retaining  Walls  with  Buttresses 97 

45.  Retaining  Walls  with  Inclined  Buttresses 98 

46.  Retaining  Walls  with  Relieving  Arches 99 

47.  Determination  of  Width  of  Base  by  Analytical  Methods   .         .     100 


CONTENTS  ix 

CHAPTER  V 
DAMS 

SECTION  PAGE 

48.  Kinds  of  Dams     .        .        .        .  '     .        .        .        .        .        .  105 

49.  Direction  of  the  Water  Pressure          .         .         .         ...  \     .  106 

50.  Amount  of  the  Pressure .106 

51.  Point  of  Application  of  the  Pressure 108 

52.  Theoretical  Profiles  for  Dams      .         .        .        .        .    '    .        .  Ill 

53.  Triangular  Profile       ...     .         -,   .• '  \        .        .        .        .        .  Ill 

54.  Trapezoidal  Profile       ,        -.        ......        .        .113 

55.  Pentagonal  Profile        .        .        .         ••       •        •        •        •        •  115 

56.  Dimensions  of  Trapezoidal  and  Pentagonal  Dams  of  Various 

Heights 118 

57.  Practical  Cross-sections        .   .     . 121 

58.  Submerged  Dams  :  Ogee  Profile  . 121 

59.  High  Dams  ...........  123 

60.  Crugnola's  Section       .        ...        .        .        .        .        .124 

61.  Krantz's  Section  ........        V        .         .  125 

62.  Author's  Section  .                          .......  125 


OF  THE 

UNIVERSITY 

OF 


EAKTH   SLOPES,    RETAINING   WALLS, 
AND   tiAMS 

CHAPTER   I 

THE   STABILITY   OF   EARTH   SLOPES 

1.  Measurement  of  Slopes.  —  When  a  trench  is  to  be  cut 
in  soil  for  temporary  use,  the  sides  of  the  excavation  are 
kept  vertical  by  sheeting  or  bracing.  But  when  the  trench 
is  to  remain  as  a  permanent  piece  of  work,  as  in  the  con- 
struction of  roads,  railroads,  canals,  etc.,  artificial  means  for 
holding  up  the  sides  are  usually  out  of  the  question,  and  to 
prevent  the  fall  of  the  sides  of  the  trench  they  are  cut  with 
a  slope  sufficient  to  enable  the  earth  to  maintain  itself  in 
place  without  support. 

The  slope  of  a  cut  or  embankment  is  represented  by  the 
ratio  of  the  base  to  the  height  of  the  right  triangle  formed 
by  the  slope  and  a  pair  of  horizontal  and  vertical  lines. 
Thus,  a  slope  of  1  to  1  indicates  that  the  base  of  the  slope 
is  equal  to  the  height;  the  slope  of  1J  to  1  that  the  base 
is  one  and  a  half  times  the  height ;  2  to  1  that  the  base  is 
double  the  height ;  and  so  on.  In  Fig.  1,  AB  represents  the 
base  and  BO  the  height  of  a  cut;  its  slope  would  be  indi- 
cated by  the  fraction  — -,  or,  if  BO=  1,  simply  by  AB:  1. 
BC 

It  is  not  possible  to  give  a  general  rule  for  the  slope  to 
which  trenches  should  be  cut,  since  in  each  individual  case 

1 


2         EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 

there  are  so  many  particular  conditions  to  be  taken  into  con- 
sideration. In  important  works  the  slopes  of  the  trenches 
should  be  fixed,  after  careful  examination  with 
regard  to  all  the  local  conditions.  Only  in  minor 
works  may  simple  practical  rules  be  followed. 
Such  are:  the  slope  of  1-|  to  1  is  usually  adopted 
for  trenches  cut  through  loose  soils ;  1  to  1  for  earth 
of  ordinary  consistency  ;  1  to  4  for  soft  rock  ;  and  1 
to  10  for  hard  and  compact  rock. 

2.  Equilibrium  of  a  Slope.  —  In  general  it  may  be  said  that 
the  slope  of  any  earth  embankment  is  the  result  of  three 
different  forces  :  (1)  the  weight  of  the  material  of  the  slope; 
(2)  the  friction  between  its  particles ;  and  (3)  the  cohesive 
forces   in    the    material,    which    keep   together   its    various 
particles. 

3.  Weight    of    Soils.  —  In    the    excavation    of    trenches 
through  loose  soils,  it  is  the  weight  of  the  material  that 
tends  to  cause  the    collapse  of  the    sides.      To  determine 
the  proper  slope  to  be  used,  therefore,  it  is  essential  to  know 
the  weight  of  the  material  concerned. 

The  weight  per  cubic  foot,  or  specific  weight  of  earth  or 
of  any  other  substance,  is  obtained  by  weighing  a  mass 
of  known  dimensions  and  dividing  by  its  volume.  But 
earth  when  removed  from  its  natural  bed  increases  in 
volume,  so  that  to  correctly  determine  its  specific  weight 
it  is  necessary  to  measure  very  accurately  the  dimen- 
sions of  the  specimen  mass  of  earth  before  excavation. 
Further,  since  moisture  increases  the  weight  of  earth, 
the  degree  of  its  moisture  content  at  the  time  the 
weight  is  determined  should  also  be  observed  and 
recorded. 


THE   STABILITY   OF   EARTH   SLOPES 


The  following  table  gives  the  weight  of  some  soils  under 
different  conditions  of  moisture  content: 


QUALITY  o 

F  SOIL 

WEIGHT  OF  1  Cu.  FT.  IN  LBS. 

dry 

....        72     to     80 

Common  Loam 

^XJ 

moist      .... 

....        70     «     76 
....        70      "     76 

full  of  water  .     . 
!drv 

....        104   "    112 
....          90    "    106 

Sand 
Gravel 

very  moist      .     . 

....        118    "    129 
....          90    "    106 

4.  Internal  Friction  of  Soils.  —  When  a  mass  of  soil  is 
thrown  into  a  pile,  it  will  dispose  itself  with  a  certain  maxi- 
mum slope  of  surface,  called  the  natural  slope.  This  slope 
is  characteristic  for  any  particular  material  in  a  given  con- 
dition, but  varies  with  the  nature  of  the  soil  and  the  amount 
of  moisture  which  it  contains. 

In  Fig.  2,  let  D  be  one  of  the  particles  of  the  earth  on  the 
surface  of  the  slope  after  the  pile  has  assumed  its  natural 
slope  at  an  angle  <£  with  the  horizontal.  Let  W  be  the 
weight  of  the  particle  D.  Resolve  the 
weight  into  its  two  components  N  and 
F  respectively  normal  to  and  parallel  to 
the  natural  slope  AB.  The  force  N 
presses  the  particle  D  against  the  rest  of 
the  soil,  while  the  force  F  will  have  a  ten- 
dency to  push  it  down  the  slope.  The  action  of  the  force  F 
is  resisted  by  the  friction  between  particle  D  and  the  rest  of 
the  soil,  under  the  pressure  N.  Since  friction  does  not  depend 
upon  the  magnitude  of  the  surface  of  contact,  but  only  upon 


FIG.  2. 


4         EARTH   SLOPES,   RETAINING  WALLS,   AND   DAMS 

the  total  pressure,  its  relation  to  this  pressure  being  expressed 
by  a  constant  coefficient  of  friction/,  we  may  conclude  that 
the  following  relation  is  true: 

F=fN.  (1) 

But  from  the  resolution  of  forces  we  know  that 


and  N=  TTcosc/>. 

Substituting  these  values  in  equation  (1)  gives 

TFsin£=/JTcos£; 
or,  in  other  terms, 

f  _  s^n  $  _ 
cos  <f> 

or  the  coefficient  of  friction  is  equal  to  the  trigonometric 
tangent  of  the  angle  of  natural  slope  of  the  earth. 

The  angle  (f>  is  called  the  angle  of  friction  and  is  used  to 
indicate  the  natural  slope  of  the  soils. 

In  practice  the  coefficient  of  friction  of  the  soils  is  obtained 
by  piling  them  up  so  that  they  may  dispose  themselves  ac- 
cording to  their  natural  slope  without  cohesive  action,  and 
observing  the  angle  of  this  slope,  given  by  the  ratio^  of  the 
horizontal  distance  between  any  two  points  taken  along  the 
line  of  the  natural  slope,  to  their  difference  of  level.  This 

ratio  is  -  or  cot  <f>. 

J 

In  the  following  table  are  given  the  angles  of  natural 
repose,  the  coefficients  of  friction,  and  the  slopes  of  various 
soils. 


THE   STABILITY   OF   EARTH   SLOPES 


QUALITY 

OF  SOIL 

ANGLE  OF  REPOSE 

COEFFICIENT 
OF  FRICTION 

SLOPE  OF 
REPOSE 

dry 

35° 

0.7002 

1.4281 

Sand 

moist 

40° 

0.8391 

1.1918 

very  wet 

30° 

0.5773 

1.7320 

Siliceous    fdry 

39° 

0.8098 

1.2349 

soils  L  moist 

44° 

0.9657 

1.0355 

Vegetable  j  dry 
soil    L  moist 

40° 
41° 

0.8391 
0.8693 

1.1918 
1.1504 

Clayey      /dry 

42° 

0.9004 

1.1106 

soil    [moist 

44° 

0.9657 

1.0355 

Gravel      {r°und 
[sharp 

30° 
40° 

0.5773 
0.8391 

1.7320 
1.1918 

From  this  table  it  may  be  seen  that  in  the  same  material 
the  value  of  <£>  changes  with  the  amount  of  moisture  that  the 
soil  contains.  Up  to  a  certain  limit,  moisture  tends  to  in- 
crease the  value  of  (/>,  while  a  larger  quantity  of  moisture 
tends  to  decrease  the  angle  of  repose.  In  fact,  when  a  fine 
soil  contains  a  large  quantity  of  water,  its  particles  run  on 
each  other,  and  the  angle  of  friction  may  decrease  to  very 
near  zero,  as  in  silts  and  muds,  which  flow  almost  like  water 
when  sufficiently  wet. 

5.  Cohesion  in  Soil.  — In  a  firm  bank  of  earth  the  force  of 
cohesion  holds  together  the  particles  of  the  material,  so  that 
they  oppose  a  certain  resistance  to  a  force  tending  to  sepa- 
rate them  from  one  another.  Generally  speaking,  the  force 
of  cohesion  can  be  considered  as  the  resistance  offered  by 
the  earth  when  being  cut. 

This  cohesive  force  varies  with  the  quality  of  the  earth ; 
it  is  very  small  in  sand  and  siliceous  soils,  and  greater  in 
clay  and  similar  materials.  When  earth  is  removed  from  its 
natural  bed,  its  cohesion  is  entirely  destroyed  ;  when  the  soil 
is  piled  up  again,  it  will  regain  its  lost  cohesion  only  in  a 


6         EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 

small  degree  unless  well  rammed  and  wetted.  Consequently 
the  cohesive  force  of  a  soil  should  be  considered  to  be  active 
only  when  the  soil  is  in  its  natural  position,  whereas  the 
force  of  friction  always  exists,  whether  the  soil  be  in  its 
original  position  or  in  an  artificial  embankment. 

On  account  of  the  cohesive  force  existing  in  soils  in  their 
natural  position,  the  slopes  of  trenches  can  be  left  more 
nearly  vertical  than  those  of  embankments,  since  in  con- 
structing embankments  the  soil  has  been  removed  from  its 
natural  bed  and  consequently  the  force  of  cohesion  destroyed. 
The  slope  of  a  cut  is  sustained  by  the  combined  action  of  the 
forces  of  cohesion  and  friction,  while  the  slope  of  an  embank- 
ment is  sustained  only  by  the  force  of  internal  friction.  It 
is  necessary,  however,  to  remember  that  little  reliance  should 
be  placed  in  the  force  of  cohesion  in  calculating  the  slopes  of 
cuts,  since  atmospheric  influences  tend  greatly  to  alter  it. 

In  opening  a  trench  through  loose  soil,  the  sides  of  the 
excavation  on  account  of  the  cohesion  of  the  material  can 
be  left  vertical  up  to  a  certain  height.  This  height  depends 
upon  the  quality  of  the  soil.  For  each  soil  there  is  a  maxi- 
mum height  at  which  it  can  remain  vertical  when  cut ;  on 
an  attempt  to  increase  the  depth  of  the  cut  beyond  this,  the 
sides  of  the  trench  will  collapse.  It  is  usual,  therefore,  to 
express  the  cohesive  force  of  a  soil  in  terms  of  the  maximum 
height  at  which  it  can  remain  vertical  when  cut.  Thus  it  is 
said  that  a  soil  stands  vertically  to  a  height  of  4.75  ft.  ;  this 
means  that  if  it  is  attempted  to  increase  the  vertical  cut  to 
say  5J  ft.,  the  sides  will  collapse. 

The  value  of  the  cohesive  force  of  earths  is  generally  ex- 
pressed by  the  force  which  it  is  necessary  to  apply  in  order 
to  destroy  this  force  per  unit  of  surface.  For  the  sake  of 
simplicity  in  the  calculations  the  force  of  cohesion  is  expressed 


THE   STABILITY   OF  EARTH   SLOPES  7 

in  terms  of  the  specific  weight  of  the  earth  multiplied  by  the 
coefficient  of  cohesion.  In  other  words,  the  coefficient  of 
cohesion  of  a  soil  is  the  ratio  of  the  cohesion  per  unit  of  sur- 
face to  the  weight  per  unit  of  volume.  Calling  this  ratio  K, 
the  force  of  cohesion  per  unit  of  surface  will  be  expressed  by 


where  y  is  the  weight  of  the  unit  of  volume  of  the  soil 
considered. 

In  the  following  table  are  given  the  values  of  the  force  of 
cohesion  in  some  soils  as  deduced  from  recent  experiments: 


QUALITY  OF  SOIL 

COHESION  IN  LBS.  PER  SQ. 

FT. 

Ordinary   earth     4 
(  moist 

Clayey  soils            •!        .  . 
{  moist 

110.8 
114.6 

107.3 
190.8 

GRAPHICAL  DETERMINATION  OF  COHESION 

6.  The  cohesive  force  in  a  bank  of  earth  can  be  easily 
determined  by  graphical  methods.  Let  a  bank  of  earth  be 
represented  as  in  Fig.  3  by 
its  profile  CBA  G- ;  and  let 
it  be  assumed  that: 

1.  The  mass  of  earth  is 
homogeneous  throughout; 

2.  The     earth     contains 
natural  moisture; 

3.  The  force  of  cohesion 
is  uniform  throughout  the 
mass  ; 

4.  The  stratification  of  the  earth  in  the  embankment  does 
not  affect  the  pressure  due  to  the  weight  of  the  material; 


FIG.  3. 


8         EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 

5.  The  portion  of  the  earth  which  tends  to  separate  from 
the  embankment  will  slide  along  a  plane  surface. 

Properly  speaking  the  last  assumption  is  not  correct,  be- 
cause the  sliding  surface  is  generally  cylindrical,  being 
formed  by  a  straight  horizontal  line  as  generatrix  moving 
along  a  cycloid  as  directrix.  But  in  order  to  facilitate  the 
calculations  it  is  here  supposed  that  the  sliding  surface  is  a 
plane.  The  results  obtained  in  the  use  of  this  assumption 
are  very  close  to  those  obtained  by  the  theoretically  exact 
method.  Besides,  the  assumption  is  on  the  side  of  safety,  as 
it  neglects  some  of  the  resisting  forces,  while  the  destructive 
forces  are  all  taken  into  consideration. 

Then  ABCG-  being  a  bank  of  earth  whose  face  has  the 
slope  AB,  and  <j>  being  the  angle  of  natural  repose  of  the 
material:  since  the  slope  AB  makes  with  the  horizontal  line 
A  0  an  angle  greater  than  <£,  the  earth  must  be  endowed  with 
cohesion,  for  otherwise  the  slope  AB  would  coincide  with  the 
slope  of  repose.  In  order  to  compute  the  value  of  this  force 
of  cohesion  which  keeps  in  equilibrium  the  portion  of  the 
earth  between  the  slope  AB  and  the  slope  of  repose  AD,  it 
is  supposed  that  the  upper  part  of  the  mass  of  earth  has 
a  tendency  to  slide  down  along  a  plane  represented  in  the 
figure  by  the  line  AGr\  then  the  prism  AEG-  will  be  under 
the  action  of  four  forces  which  are  in  equilibrium,  as  follows: 

1.  The  weight  TFof  the  prism  AB&; 

2.  The  reaction  N  that  the  lower  mass  exerts  upon  the 
prism  ABGr,    in  a  direction  perpendicular   to  the    plane  of 
sliding; 

3.  The  friction  F  along  the  plane  AG-,  depending  upon 
the  pressure  Nf  equal  and  opposite  to  the  reaction  N.     This 
force  of  friction  opposes  the  descending   movement  of  the 
prism ; 


THE   STABILITY   OF   EARTH   SLOPES  9 

4.  The  resistance  offered  by  the  cohesion  0  acting  in  the 
direction  of  AGr  which  tends  to  oppose  the  sliding  of  the 
prism. 

-  Draw  the  force  polygon  :  along  a  vertical  line  lay  off  a 
segment  ab  equal  to  the  weight  TFof  the  sliding  prism  ;  from 
a  draw  the  line  ac  parallel  to  AGr  and  from  b  draw  a  normal 
to  AGr.  Since  the  forces  are  in  equilibrium  the  polygon  abc 
will  be  a  closed  one.  The  line  ac  represents  the  forces  which 
prevent  the  sliding  of  the  prism,  which  are  the  friction  and 
the  cohesion.  Therefore  ac  must  be  equal  to  F  +  C.  Sup- 
pose d  to  be  the  dividing  point  between  these  two  quantities, 
so  that  cd  represents  the  friction  and  da  the  cohesion  ;  join  b 
and  d.  The  forces  F  and  N  are  perpendicular  to  each  other 
and  their  resultant  Q  makes  with  N  an  angle  equal  to  the 
angle  of  repose  fa  or  in  other  words  the  angle  cbd  must  be 
drawn  equal  to  <£. 

Let  the  length  of  the  embankment  be  equal  to  unity,  i.e. 
let  the  depth  of  the  prism  normal  to  the  plane  of  the  paper 
be  equal  to  1.  From  the  point  B  drop  the  perpendicular 
BE  on  A  G-  ;  then  the  weight  W  of  the  prism  will  be  given  by 


where  7  is  the  weight  of  the  material  per  unit  of  volume. 

To  represent  the  forces  by  straight  lines,  it  is  necessary  to 
assume  a  scale.  Let  in  this  case  the  scale  of  reduction  for 
the  forces  be  J  yA  Gr  so  that  in  the  construction  of  the  force 
polygon  we  shall  have 


Now,  AD  being  the  natural  slope,  BD  a  perpendicular 
from  B  upon  AD,  and  EF  an  horizontal  line  drawn  from 
E,  the  triangle  BEF  will  be  equal  to  the  triangle  dba  of  the 


10      EARTH   SLOPES,    RETAINING   WALLS,   AND   DAMS 

force  polygon.  For,  BE  =  W  =  ab,  ^.BEF  =  Z  lac  having 
the  sides  respectively  perpendicular,  A  /.  EBF  =  Z  abd  (be- 
cause cba  =  G- AC  having  their  sides  respectively  perpendic- 
ular, and  Z.cbd  =  DAO  =  <£,  hence  /-dba  =  GAD  =  EBF}, 
which  makes  the  triangles  equal,  having  one  side  and  two 
angles  equal.  Therefore  da  =  EF.  But  in  the  force  polygon 

ca  =  cd  +  da  =  F  +  C, 

and  since  the  construction  made  cd  =  F,  we  have  by  sub- 
traction 

da  =  O,  and  consequently  O  =  EF. 

Reducing  from  the  assumed  scale,  we  find  that  the  force  of 
cohesion,  or  the  resistance  due  to  the  cohesion,  is 

x  EF. 


Since  by  assumption  the  force  of  cohesion  is  uniformly  dis- 
tributed, it  acts  at  all  points  along  the  surface  AGr  and  its 
value  per  unit  of  surface  will  be 

O 


By  definition,  the  coefficient  of  cohesion  is  the  quantity  by 
which  it  is  necessary  to  multiply  the  specific  weight  of  the 
material  in  order  to  obtain  the  force  of  cohesion  per  unit  of 
surface.  But  y  being  the  unit  of  weight  of  the  material, 
the  coefficient  of  cohesion  K  is  K  =  ^  EF. 

It  is  evident  that  the  coefficient  of  cohesion  K  is  given  in 
terms  of  the  line  EF,  which  varies  with  the  different  posi- 
tions assumed  by  the  probable  planes  of  sliding.  Conse- 
quently it  will  be  interesting  to  determine  what  direction  of 
A  G-  gives  the  greatest  value  of  K. 

Suppose  the  plane  of  sliding  AGr  to  turn  around  the  point 
A,  Fig.  4.  The  line  EF  is  easily  determined  for  each  new 


THE   STABILITY   OF   EARTH   SLOPES  11 

position  of  AG.  The  angle  BE  A  will  always  be  a  right 
angle ;  therefore  the  point  E  will  describe  a  circular  arc 
described  on  AB  as  diameter.  This  semicircle  passes  through 
D,  as  Z  BDA  is  a  right  angle.  Since  BD  is  a  fixed  line, 
because  AD,  the  slope  of  repose,  is  fixed,  it  is  evident  that 
EF  will  reach  its  greatest  value  when  E  falls  in  the  middle 
of  the  arc  BD,  or  at  E' .  Consequently  we  have  :  maximum 
value  of  K—  |  E'F' ,  and  this  is  obtained  when  A  G-  falls  upon 
the  line  AG-'  bisecting  the  angle  BAD. 

This   means    that    the    cohesion    re- 
quired to  hold  the  mass  from  sliding 
reaches   its    greatest    value    when   the 
probable  plane  of  sliding  bisects 
the  angle  between  the  slope  AB 
and  the  plane  of  natural  repose 
of  the  material  AD.     The  plane         A   "'—""  C 

A  G  is  therefore  to  be  considered  FlG*  4* 

as  the  most  probable  plane  of  sliding,  the  plane  along  which 
the  prism  of  earth  which  tends  to  split  off  from  the  embank- 
ment would  separate.  In  the  considered  embankment,  if 
the  force  of  cohesion  of  the  earth  is  greater  than  ^E'F'<y, 
the  earth  can  be  maintained  under  the  slope  AB,  and 
^<yE'F'  will  then  represent  the  force  of  cohesion  actually 
developed  along  the  probable  plane  of  rupture. 

In  Fig.  4,  draw  the  line  BE'  perpendicular  to  AG-'  and 
produce  it  to  E"  on  the  plane  of  natural  repose;  from  this 
point  JE"  draw  the  horizontal  line  E"F". 

AE"  =  AB  and  E"F"  =  2  E'F'. 

For,  since  E'  is  the  middle  point  of  the  arc  BD,  the  angle 
BE' A  is  a  right  angle,  as  is  also  the  angle  AE'E"-,  and 
AG  being  the  bisector  of  the  angle  BAE",  the  triangles 


12      EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 

BAE1  and  EAE"  are  equal,  which  gives  AB  =  AE",  and 
BE'=E'E",  or  BE"  =  2  BE1.  The  triangles  BE'F  and 
BE"Fn  are  similar,  having  one  angle  common  and  the 
opposite  sides  parallel.  Since  the  side  BE"  is  equal  to 
twice  BE',  and  in  similar  triangles  all  sides  are  in  the  same 
proportion,  we  find  that  E"F"'*=2E'F.  Consequently  the 
maximum  value  of  the  coefficient  of  cohesion  K  \v\\\  be 


The  graphical  construction  for  finding  the  coefficient  of 
cohesion  may  be  described  as  follows  :  from  B  draw  a 
perpendicular  to  AD,  the  line  of  natural  slope;  then  make 
AE"  =AB  and  from  E»  draw  E"F"  horizontally  to  intersect 
BD  in  F".  The  coefficient  of  cohesion  Ki§ 

K=  \  E"F". 

In  the  right-angled  triangle  F"DE"  the  angle  at  E"  =  0, 
because  CAE"  and  AE"F"  are  alternative  interior  angles. 
Therefore  DJE"  =  E"F"  cos  <t>  and,  since  K=\E"F",  or 
E"F"  =  4  K,  we  find, 


Thus,  the  coefficient  of  cohesion  K  can  be  expressed  in  terms 

of  DE", 

DE" 


THE  PARABOLA  OF  COHESION;  PRACTICAL  APPLICATIONS 

7.  The  preceding  has  dealt  with  the  fact  that  an  earth 
bank  is  in  equilibrium  so  long  as  its  side  does  not  exceed  a 
certain  slope.  We  now  will  be  concerned  with  the  further 
observation  that  the  greater  the  height  of  a  bank  or  pile, 
the  flatter  will  be  the  limiting  slope.  This  observation  may 


THE   STABILITY   OF   EARTH   SLOPES  13 

be  put  more  precisely  by  saying  that,  in  a  general  way,  the 
limiting  slope  is  inversely  proportional  to  the  height  of  the 
bank. 

It  will  be  interesting  to  study  the  changing  position  of 
the  top  of  the  embankment  B,  as  the  embankment  increases 
in  height,  supposing  it  to  be  maintained  at  the  maximum 
slope.  This  will  afford  a  simple  way  of  determining  the 
proper  slope  to  be  given  to  a  bank  which  is  to  remain  in 
equilibrium  under  the  combined  forces  of  friction  and  cohe- 
sion. 

In  a  particular  bank  of  earth 
the  forces  of  friction  and  cohe- 
sion have  a  constant  maximum 
value,  i.e.  the  value  of  <f>  and 
K  are  constant  throughout  the 
mass. 

From  B,  Fig.  5,  draw  BD 
perpendicular  to  the  line  of  FlG'  5> 

natural  slope  and  take  AE—AB\  produce  the  line  AE  and 
lay  off  a  segment  AM  =  DE  ;  at  M  erect  a  perpendicular 
to  the  ME  and  from  B  draw  the  line  BN  parallel  to  the 
natural  slope  AE.  It  will  result,  then,  that 


and 


which  is  a  constant  quantity  because  both  $  and  K  are 
constant.  The  straight  line  MN,  therefore,  has  a  fixed 
position.  Since  the  line  BN  is  always  equal  to  AB,  it  fol- 
lows that  the  point  B  is  equidistant  from  the  fixed  point  A 
and  the  fixed  line  MN,  no  matter  what  height  be  assigned 
to  the  bank.  This  equality,  it  happens,  is  the  character- 


14      EARTH   SLOPES,   RETAINING   WALLS,   AND  DAMS 


<£ 


FIG.  6. 


istic  of  a  parabola.  The  locus  of  point  B,  the  top  of 
the  bank  of  maximum  slope,  is  therefore  a  parabola  whose 
focus  is  at  A  and  whose  directrix  is  the  line  MN\  the  slope 
of  repose,  AD,  is  the  axis  of  the  parabola. 

The  parabola  of  cohesion  gives,  for  any  given  bank  of 
earth,  the  various  slopes  of  equilibrium  corresponding  to 

the  different  heights.  Thus,  for 
instance,  in  Fig.  6,  draw  from  A 
a  perpendicular  to  the  line  AC; 
the  point  of  intersection  with  the 
parabola  will  determine  the  height 
up  to  which  a  vertical  face  will 
remain  in  equilibrium.  Again, 
the  parabola  of  cohesion  indicates 
that  for  heights  less  than  this,  the 
earth  will  stand  even  inclined  for- 
ward from  the  perpendicular,  a  phenomenon  which  is  illus- 
trated along  the  edges  of  rivers  and  creeks,  where  the  bank 
is  undercut  by  the  stream. 

The  following  examples  of  the  use  of  the  parabola  of 
cohesion  may  be  serviceable  to  the  student : 

1.  It  is  desired  to  know  the  slope  which  shall  be  given 
to  a  bank  of  earth  25  ft.  high,  when  it  is  known  that  the 
material  can  stand  with  a  vertical  face  up  to  a  height  of 
7ft. 

From  J.,  the  foot  of  the  embankment,  Fig.  6,  draw  a 
vertical  line  and  make  AB  =  1  ft.,  then  draw  the  line  AM 
on  the  slope  of  repose,  making  with  the  horizontal  an  angle 
equal  to  <f>.  On  AM  lay  off  a  segment  AE  =  AB,  from  B 
draw  the  line  BD  perpendicular  to  AM,  from  E  the  hori- 
zontal line  UF;  then  \EF  =  K,  the  coefficient  of  cohesion. 
Produce  AM  and  lay  off  a  segment  AA'  =  DE  and  describe 


Rsrnrl 


THE   STABILITY   OF   EARTH   SLOPES  15 

a  parabola  with  vertex  halfway  between  A  and  A,  and  with 
focus  at  A.  Then  drawing  a  horizontal  line  25  ft.  above 
the  base,  where  this  intersects  the  parabola,  we  find  the  top 
edge  of  the  bank  N.  Connect  N  with  A ;  the  slope  AN 
is  the  maximum  slope  of  equilibrium  corresponding  to  the 
height  of  25  ft. 

2.  If  instead   of   the   vertical   height   we   are    given   K, 
the  coefficient  of  cohesion  of  the  material,  the   solution  is 
almost   the    same,   the   only  difference   being  that  the  pa- 
rabola is  located  by  making   the  focal  distance    A  =  A  = 
4  K  cos  (/>. 

3.  The  Stepped    Slope.  —  In    high    embankments   instead 
of   using  a  single  slope  it   is  usually  preferred  to  cut  the 
earth  in  successive  small  slopes 

alternate  with  benches.  The 
parabola  of  cohesion  gives  an 
elegant  solution  of  the  problem 
of  proportioning  the  successive 
slopes. 

Let   h,   Fig.    7,   be    the    total 
height  of  the  embankment ;  no, 
op,  pq,   the  height  of    the  suc- 
cessive equal  steps ;    and  <f>  the  FIG.  7. 
angle  of  natural  repose  of  the  material. 

Draw  the  parabola  of  cohesion  as  in  any  other  case  (the 
coefficient  of  cohesion  or  equivalent  information  being  given, 
as  in  the  preceding  examples).  From  the  points  o,  p,  and  q, 
draw  horizontal  lines,  intersecting  the  parabola  at  0,  P,  Q, 
respectively.  Connect  A  with  Q,  then  the  line  AQ  repre- 
sents the  slope  of  equilibrium  for  the  total  height  h  of  the 
embankment.  Connect  also  A  with  0  and  P.  The  lines 
A  0  and  AP  will  represent  the  slopes  corresponding  to  one 


16      EARTH   SLOPES,   RETAINING   WALLS,   AND  DAMS 

and  two  steps  respectively.  Produce  the  line  oO  to  meet 
at  B  the  slope  AQ  and  lay  off  a  segment  BD  equal  to  the 
given  width  of  the  bench  or  berm.  From  D  draw  DE 
parallel  to  the  slope  AP,  intersecting  the  line  pP  produced 
at  E,  and  again  lay  off  a  horizontal  segment  EF  equal  to  the 
required  width  of  berm.  Finally  from  F  draw  a  parallel  to 
the  slope  AO  until  it  meets  at  Cr  the  line  AQ.  The  broken 
line  ABDEFG-  will  be  the  required  profile  of  the  cut. 

4.  The  Slope  of  Equal  Stability.  —  The  parabola  of 
cohesion  also  affords  an  easy  solution  of  the  problem  of 
determining  in  a  graphical  way  the  slope  of  equilibrium 

of  equal  stability  for  a  bank  of 
earth.  The  slope  of  equal  sta- 
bility differs  from  the  slope  of 
equilibrium  corresponding  to  a 
certain  height;  while  the  latter 
has  been  considered  as  a  straight 
line,  the  former  is  a  curve.  It 
is  observed  in  trenches  that  have 
originally  been  cut  vertical,  and 
FlG<  8-  then  were  exposed  to  atmos- 

pheric influences  for  a  long  time,  that  the  face  assumes  a 
curved  outline  ;  this  is  the  slope  or  surface  of  equal  stability. 
Suppose  that  in  Fig.  7  the  total  height  h  of  the  embank- 
ment was  equal  to  24  ft.,  and  each  one  of  the  various 
sections  no,  op,  pq  equal  to  8  f  t. ;  A  Q  then  is  the  proper 
slope  for  the  total  height  of  24  ft.,  AP  for  the  height  of 
16  ft.,  and  AO  for  the  height  of  8  ft.  The  lowest  step  was 
given  the  slope  proper  to  the  height  of  24  ft.,  the  total 
height  of  the  embankment,  and  thus  was  drawn  the  line  AB. 
The  remaining  height  from  B  to  the  top  of  the  embankment 


THE   STABILITY   OF  EARTH   SLOPES 


17 


is  16  ft.,  and  for  the  second  section  was  used  the  slope  proper 
to  the  height  of  16  ft.,  drawing  DE  parallel  to  AP.  For 
the  upper  portion  was  used  the  slope  proper  to  an  embank- 
ment 8  ft.  high  by  drawing  F&  parallel  to  A  0. 

If  it  were  not  for  the  berms,  the  various  slopes  would  have 
formed  a  broken  line,  as  indicated  in  Fig.  8.  It  is  evi- 
dent that  the  smaller  the  steps  into  which  the  total  height 
of  the  embankment  is  divided,  the  more  closely  the  broken 
line  approximates  to  a  curve.  A  division  into  seven  steps 
is  shown  in  Fig.  9.  When  the  points  are  taken  at  an  in- 
finitesimally  small  distance  from  one  another,  the  resultant 
slope  of  equilibrium,  a  slope  / 

of  equal  stability  at  all  heights, 
becomes  a  smooth  curve. 

Once  the  slope  proper  to  a 
given  height  of  a  bank  of  a 
given  quality  of  soil  having 
both  friction  and  cohesion  is 
known,  it  is  then  an  easy  mat- 
ter to  determine  the  safe  slope 
for  a  bank  in  the  same  soil  FIG.  9. 

of  any  different  height.  Evidently,  also,  if  it  is  desired  to 
determine  the  slope  which  will  call  into  play  only  |,  ^,  ^  of 
the  total  force  of  cohesion,  so  that  the  bank  is  designed  with 
a  factor  of  safety  of  2,  3,  4,  etc.,  in  respect  to  its  force  of  cohe- 
sion, it  is  only  necessary  to  find  the  slope  of  equilibrium  or 
the  curve  of  equal  stability  for  a  coefficient  of  cohesion 
taken  at  -|,  J,  ^  of  the  actual  value.  For  this  purpose  the 
parabola  of  cohesion  is  employed  in  the  same  manner  as  in  the 
preceding  cases;  the  focal  distance,  however,  instead  of 
being  taken  equal  to  4  K  cos  <£,  is  made  J,  J,  \  of  this  value. 


18     EARTH  SLOPES,  RETAINING  WALLS,  AND  DAMS 

ANALYTICAL  CALCULATION  OF  COHESION 

8.  Slope  without  Surcharge.  —  The  influence  of  the  cohe- 
sion can  be  calculated  also  by  analytical  methods.  Consider 
an  embankment  under  the  slope  AB,  and  denote  by 

<f>  the  angle  of  repose, 

/3  the  angle  included  between  the  slope  AB  and  the  plane 
of  natural  repose  represented  by  the  line  AD, 

I  the  length  of  the  slope  AB, 

h  the  vertical  height  of  the  embankment,  and 

^Tthe  coefficient  of  cohesion. 

From  the  preceding  section  we  know  that,  in  Fig.  5, 


but  DE  =  AKcos  <f>  and  AD  =  AB  cos  fi  =  I  cos  j3. 

Then  the  length  of  the  slope  of  the  embankment  will  be 

1=1  cos  /3  +  A  7Tcos  <j>, 
from  which  is  deduced  the  value  of  the  coefficient  of  cohesion, 


4  cos<£ 

R  R 

but  cos  /3  =  cos2  ^  —  sin2  £-  , 


and  I  -Zcos  j3  =  I  -?cos      -  - 

'—  — 

R  R 

and  since  I  —  &os2  £•  =?sin2  ^, 

—  — 


R 
we  shall  have  I  — 7cos  ft  =  2Zsin2  ^ 


THE   STABILITY   OF   EARTH   SLOPES  19 

Substituting  this  value  in  the  formula  (V),  we  have 


sin2| 
K=l 


2  cos<£ 

Thus  the  coefficient  of  cohesion  K  is  given  by  the  length 
of  the  slope  of  the  embankment  multiplied  by  the  square  of 
the  sine  of  half  the  angle  between  the  slope  and  the  plane 
of  natural  repose,  divided  by  twice  the  cosine  of  the  angle  of 
repose  0. 

The  coefficient  of  cohesion  can  be  found  also  in  terms  of 
the  height  of  the  embankment.  From  Fig.  5  it  is  easily 
seen  that  Ji  =  I  cos  d,  where 

d=90-(/3  +  <£);   then  cos  d  =  sin 

and  1= 

sin 

Substituting  this  value  in  equation  (5),  we  shall  have  the 
coefficient  of  cohesion  K  in  terms  of  the  height  of  the  em- 
bankment and  both  the  angles  <£  and  (3  :  • 


00 


Sin2 


The  equations  ((?)  and  (c?)  involve  only  the  four  quan- 
tities jfif,  A,  /3,  and  <£,  so  that  when  any  three  of  them  are 
given  the  fourth  may  be  calculated. 

The  force  of  cohesion  is  often  given  in  terms  of  the  maxi- 
mum height  at  which  the  earth  will  stand  with  vertical  face. 


20      EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 


For  a  vertical  face,  £  +  £  =  90°.  Call  hr  the  value  of  Z,  or 
the  height  at  which  the  embankment  remains  vertical  due  to 
the  cohesion.  Observe  that 


cos  </>  =  sin  (90  -  0)  =  2  sin          -       cos 

%  2 

Substitute  these  values  in  equation  (b)  ;  then 

90  -  <f> 

-£ 

=  -    tan    45  - 


sm 


dn/TSzT-* 

(    2 


,\         A)0  —  d>\       4 
)COS(     2     ) 

from  which  is  deduced 

A'  =  4.Zf  cot  f45-l}  =  44Ttan  (45  +  ^)-  (/) 


Putting  in  equation  (<?)  the  value  of  K  given  by  equation 
(/),  we  obtain  the  value  of  h  in  terms  of  hr  : 


cos 


This  formula  gives  the  value  of  h  when  hf,  /3,  and  <£  are 
given  ;  or  it  can  be  used  for  calculating  the  angle  /3  +  <£  to 
be  given  to  a  slope  when  h',  $,  and  h  are  known. 

9.  Surcharged  Slope.  —  Consider  a  mass  of  earth  limited 
above  by  a  surface  BD  which  makes  an  angle  a  with  the 
horizontal  line  passing  through  B.  The  embankment  will 
then  be  surcharged  by  an  additional  weight  W  per  unit  of 
surface,  and  the  limiting  height  for  a  given  slope  ft  -f  <j>  will 
be  decreased. 


THE   STABILITY   OF   EARTH   SLOPES 


21 


In   Fig.    10,  AD  represents  the  plane  of   sliding  of   the 
prism   ABD,    W  its  weight  with   surcharge    included,  and 

c 


FIG.  10. 


7  the  unit  of  weight  of  the  earth.     The  total  weight  of  the 
prism  is 


W=  7 


sin 


or  making  7  =  1  and  factoring, 


ABxBD    . 

sin 


2  W  \ 

sin  (ft  -f-  4>  —  a)/ 


But 


cos  (90  -  £  -  0)      sin 
and  substituting  this, 
in 


which  means  that  the  weight  of  the  sliding  prism  increased 

by  the  surcharge  by  a  quantity       .    **!!!,'*'  ^N  ,  the  maxi- 

A  sin  (p  -H  9  —  a) 


22       EARTH   SLOPES,   RETAINING  WALLS,   AND   DAMS 

mum  height  h  at  which  the  bank  will  stand  can  be  deduced 
from  equation  (df)  corrected  by  this  factor,  thus : 

2  jfTsin  (ff  +  <ft)  cos  ft  _  ^  A 
sin2  — 


from  which  is  deduced  the  value  of  /*, 

sin  (/3-f-  c/>)  cos  <£      2  TTsin 


sin 
sin"  '- 


.2 


If  an  earth  embankment  can  remain  vertical  to  the  height 
Ar  and  then  is  surcharged  with  an  equally  distributed 
weight  W  per  unit  of  surface,  the  height  at  which  it  will 

now  stand  vertical  is  decreased  in  the  ratio sm  (  ^  "*"  ft ' . 

sin  (/3  +  </>  -  «; 

This  factor  can  be  marked  directly  on  the  slope  by  mak- 

2  TF 

ing  BB'  = ,  which  is  obtained  by  drawing  a 

sm  (£  +  <£-<*) 

line  BlCl  parallel  to  B 0 at  a  distance  BF  from  B  equal  to 
2  W.  When  the  surface  above  BO  is  horizontal,  «  =  0  and 
BF=  2  W  gives  directly  the  decreased  height  of  the  slope. 

EARTH  SLOPES  IN  PRACTICE 

10.  In  public  works  it  is  a  common  practice  to  use  the 
slope  of  1  to  1  for  cuts  and  the  slope  of  1J  to  1  for  fills. 
The  convenience  of  such  a  general  rule  is  founded  upon  the 
fact  that  it  is  impracticable  to  calculate  the  proper  slopes 
for  the  various  cuts  and  fills  encountered  in  any  ordinary 
piece  of  work.  Since  the  character  of  the  soil  changes  con- 
tinuously along  the  line  of  the  work,  it  would  be  a  very  slow 
and  expensive  matter  to  calculate  the  proper  slopes;  and  the 


THE   STABILITY  OF   EARTH   SLOPES  23 

resulting  delays  and  the  increased  cost  of  work  would  not 
bring  any  material  benefit.  For  this  reason  short  practical 
rules  are  always  followed.  But  the  convenience  of  the 
practical  rules  should  not  mislead  the  engineer  when  he 
has  to  deal  with  a  special  problem,  as,  for  instance,  making  a 
deep  cutting  for  the  approaches  of  a  tunnel  in  railroad  work, 
or  for  a  canal.  In  these  and  similar  cases,  giving  to  the 
cuts  their  proper  slopes  will  result  in  great  savings  both  in  the 
original  cost  of  construction  and  in  the  cost  of  maintenance. 

The  slope  of  1-|-  to  1  commonly  given  to  fills  is  a  safe  slope 
in  most  cases.  The  earth  in  a  fill  stands  up  only  by  virtue 
of  the  force  of  friction,  since  the  force  of  cohesion  has  been 
destroyed  in  the  removal  of  the  soil  from  its  natural  bed. 
From  the  table  given  on  p.  5  it  is  seen  that  the  natural 
slope  of  most  of  the  common  soils  is  smaller  than  1.5.  Hence 
the  slope  of  1|-  to  1  as  generally  used  may  be  considered  safe. 
But  there  are  soils,  as  rounded  gravel,  for  instance,  that  have 
a  natural  slope  as  high  as  1.7  to  1;  consequently,  when  such 
materials  are  used  in  forming  an  embankment,  a  slope  of  If 
to  1  should  be  given  instead  of  the  usual  slope  of  1J  to  1. 

The  slope  of  1|  to  1  given  to  a  certain  embankment  may 
be  considered  a  safe  slope,  when  the  material  is  dry  or  con- 
tains only  a  small  percentage  of  water ;  but  when  the  soil 
contains  a  large  quantity  of  water  it  may  assume  a  smaller 
slope.  Thus,  in  the  same  table,  p.  5,  the  natural  slope  for 
sand  when  dry  is  given  as  1.43  to  1,  when  moist  as  1.19  to  1, 
when  very  wet,  1.73  to  1.  From  this  it  can  be  seen  that  in 
determining  the  slope  to  be  given  to  an  embankment,  chiefly 
composed  of  sand,  it  is  safer  to  give  a  slope  of  2  to  1,  instead 
of  the  usual  slope,  unless  local  conditions  insure  the  prompt 
discharge  of  water  and  make  it  practically  impossible  for 
the  embankment  to  become  saturated  with  water. 


24       EARTH   SLOPES,   RETAINING  WALLS,   AND   DAMS 

It  is  in  the  cuts  that  practical  rules  should  be  applied 
most  cautiously.  If  a  bank  of  earth  is  left  with  the  slope 
of  1  to  1,  while  the  angle  of  repose  of  all  loose  materials  is 
smaller  than  45°,  it  follows  that  the  engineer  takes  advan- 
tage of  the  cohesive  force  of  the  soil  to  maintain  the  bank. 
From  the  parabola  of  cohesion  we  learn  that  the  higher  the 
bank  of  earth,  the  smaller  is  its  slope  of  equilibrium.  It 
is  evident  that  it  would  be  more  economical  to  cut  the  sides 
of  the  bank  to  a  compound  or  curved  slope,  closely  follow- 
ing the  curve  of  equal  stability,  instead  of  giving  to  the 
total  height  of  the  embankment  a  single  slope,  since 
the  latter  involves  a  larger  amount  of  excavation.  It 
is  not  necessary  that  the  slope  of  equal  stability  be 
designed  for  the  earth  in  equilibrium;  it  may  be  designed 
with  a  certain  factor  of  safety  in  respect  to  the  cohesive 
power  of  the  soil,  this  factor  of  safety  being  assumed 
according  to  the  local  conditions,  as  will  be  discussed 
farther  on. 

Using  for  deep  trenches  side  slopes  following  the  curve  of 
equal  stability,  designed  with  a  factor  of  safety  of  2,  for 
instance,  only  one  half  the  available  force  of  cohesion  is 
brought  into  play  all  through  the  mass  of  the  bank  of  earth, 
and  consequently  also  at  the  bottom  of  the  bank.  In  the 
practical  slope  of  1  to  1  there  is  a  very  large  factor  of  safety 
in  the  upper  part  of  the  bank,  which  means  that  there  is  a 
large  amount  of  useless  excavation  done  on  the  upper  section 
of  the  cut,  while  near  the  foot  of  the  bank  the  force  of  cohe- 
sion is  taken  almost  at  the  equilibrium  point.  The  foot  of  the 
bank  is  therefore  the  most  dangerous  point,  the  one  most 
liable  to  a  collapse.  In  actual  work  it  is  the  lower  portion 
of  the  bank  that  always  collapses  first,  and  consequently  the 
smallest  slope  should  be  given  to  the  lower  part  of  the  cut. 


THE   STABILITY   OF   EARTH   SLOPES 


25 


This  is  obtained  only  by  designing  the  slope  according  to 
the  curve  of  equal  stability. 

Figure  11  represents  the  various  slopes  of  equal  stability 
and  the  practical  slope  of  1  to  1,  for  a  bank  of  earth  50  ft. 
high,  in  which  the  coefficient  of  cohesion  was  assumed  to  be 
equal  to  1.3,  and  the  angle  of  natural  repose  =  30°.  The 
line  AB  indicates  the  slope  of  1  to  1;  the  slope  of  equal 
stability  designed  with  the  value  of  cohesion  taken  at  the 
equilibrium  point  is  indicated  by  J.<7,  while  AD  and  AE 


FIG.  11. 


represent  the  slope  designed  with  factors  of  safety  2  and  3 
respectively.  Comparing  these  various  slopes  it  is  seen  that 
the  slope  of  equilibrium  forms  an  angle  of  50°  with  the  hori- 
zontal at  the  lowest  portion  of  the  cut,  while  the  slope  of 
equal  stability  designed  with  a  factor  of  safety  of  2  forms  an 
angle  of  43°  15',  and  the  slope  designed  with  a  factor  of 
safety  of  3,  an  angle  of  40°  15'. 

The  amount  of  excavation  back  of  the  vertical  for  a  bank  1 
ft.  in  length  will  be  1250  cu.  ft.  for  the  slope  of  1  to  1,  873.5 
cu.  ft.  for  the  slope  of  equilibrium,  1165.5  for  a  slope  with  fac- 


26       EARTH. SLOPES,   RETAINING  WALLS,  AND   DAMS 

tor  of  safety  2,  and  1260  cu.  ft.  for  the  slope  with  a  factor  of 
safety  3.  From  this  it  is  seen  that  the  usual  slope  of  1  to  1 
requires  almost  the  same  amount  of  excavation  as  a  slope  de- 
signed with  a  factor  of  safety  equal  to  3 ;  while  the  former 
has  an  angle  of  45°,  the  second  forms  with  the  horizontal 
an  angle  of  40°  15'.  A  curve  of  equal  stability  in  which 
the  lowest  portion  makes  an  angle  of  45°  with  the  horizontal 
would  require  only  950  cu.  ft.  of  excavation,  a  saving  of 
nearly  ^  the  total  amount. 

If  such  economy  can  be  obtained  in  cuts  of  only  50  ft.,  it 
is  easy  to  imagine  the  enormous  amount  of  excavation  which 
could  be  saved  by  using  the  curve  of  equal  stability  in  cutting 
very  deep  trenches,  as  for  instance  at  the  great  cuts  of  the 
Panama  Canal,  300  ft.  deep  and  extending  several  miles  in 
length. 

Owing  to  the  fact  that  in  deep  trenches  heterogeneous  ma- 
terial is  encountered  as  a  rule,  it  would  be  very  difficult  to 
determine  the  force  of  cohesion  of  the  various  materials  and 
integrate  the  various  slopes  of  equal  stability  in  a  single 
curve  best  suited  to  the  particular  case  under  consideration. 
But  the  engineer  should  generally  adopt  a  slope  proper  to 
the  material  endowed  with  the  smallest  cohesive  force.  In 
doing  this  he  must  not  forget  that  the  lower  strata  of  the 
bank  are  most  liable  to  collapse,  so  that  when  a  very  loose 
material  is  met  with  in  the  upper  strata,  the  cut  may  be 
made  with  a  slope  greater  than  if  the  same  material  were 
encountered  farther  down. 

Since  the  force  of  cohesion  in  any  embankment  is  greatly 
altered  by  atmospheric  influences,  it  is  safest  to  rely  only  on 
a  fraction  of  it,  or  in  other  words  to  use  the  force  of  cohesion 
with  a  certain  factor  of  safety,  2  or  3,  according  to  local  con- 
ditions. A  factor  of  safety  of  2  would  give  sufficient  secur- 


THE   STABILITY   OF   EARTH   SLOPES  27 

ity  under  the  climatic  conditions  of  this  country,  while  a 
factor  of  safety  of  at  least  3  should  be  used  in  countries,  like 
Panama,  situated  in  the  tropical  regions  where  heavy  rains 
occur  in  some  part  of  the  year  so  frequently  as  to  entirely 
permeate  the  soil. 


CHAPTER   II 

RETAINING  WALLS 

11.  Theories  of  Earth  Pressure.  —  Retaining  walls  are 
erected  for  the  purpose  of  supporting  earth  embankments  or 
the  sides  of  cuts,  at  slopes  steeper  than  would  be  in  equilib- 
rium without  artificial  support.  In  a  bank  which  is  held 
up  by  a  retaining  wall,  the  internal  forces  (friction  and 
cohesion)  are  not  sufficient  to  keep  the  face  from  caving. 
Consequently  the  bank  presses  against  the  wall  with  a  cer- 
tain intensity  of  pressure.  The  proportions  of  the  wall  will, 
of  course,  be  determined  by  the  amount  of  this  pressure,  so 
that  in  order  to  design  a  retaining  wall,  the  pressure  which 
the  structure  has  to  resist  must  first  be  calculated. 

Although  many  authors  have  elaborated  theories  of  earth 
pressure  against  walls,  all  these  theories  can  be  classified  in 
two  groups:  (1)  the  theory  of  the  sliding  prism,  and  (2)  the 
analytical  theory. 

The  theory  of  the  sliding  prism  originated  with  Vauban, 
a  general  in  the  French  army  in  1687.  General  Vauban 
assumed  that  the  natural  slope  is  constant  for  all  soils  and 
equals  45°,  and  that  the  triangular  prism  of  soil  above  the 
plane  of  repose,  sliding  on  that  plane,  causes  a  thrust  equal 
to  the  actual  thrust  against  the  retaining  wall.  He  resolved 
the  weight  of  this  sliding  prism  into  components  normal 
and  parallel  to  the  natural  slope,  and  considered  the  second 
component  as  the  real  pressure  against  the  wall. 

28 


RETAINING   WALLS  29 

Vauban's  theory  was  improved  by  Belidor  in  1729.  Beli- 
dor  considered  that  the  pressure  against  a  wall  is  not  equal 
to  the  tangential  component  of  the  sliding  prism,  as  stated 
by  Vauban,  but  is  less  than  this  component  by  a  certain 
amount  due  to  the  friction  of  the  sliding  prism  on  the  soil 
beneath.  Finally,  Captain  Coulomb  of  the  French  army, 
in  the  year  1773,  still  further  developed  the  theory  of  the 
sliding  prism.  He  resolved  the  reaction  of  the  wall  (equal 
and  opposite  to  the  pressure)  and  the  weight  of  the  sliding 
prism  into  their  components  parallel  and  normal  to  the 
plane  of  rupture.  He  made  the  difference  of  the  compo- 
nents parallel  to  the  plane  of  rupture  equal  to  the  fric- 
tional  resistance  opposed  by  the  mass  of  earth  below  the 
sliding  prism,  and  from  this  equation  deduced  the  value 
of  the  total  pressure  against  the  wall.  This  theory, 
later  perfected  by  Prony,  Fran9ois,  and  Poncelet,  is  the 
one  most  commonly  used ;  it  is  generally  known  as 
Coulomb's  theory  of  retaining  walls,  or  the  theory  of  the 
sliding  prism. 

In  the  year  1856,  Professor  W.  M.  Rankine  deduced  the 
value  of  the  pressure  of  the  earth  against  a  retaining  wall 
by  means  of  a  new  and  more  scientific  principle.  He  inves- 
tigated the  conditions  of  equilibrium  of  an  interior  element 
in  a  homogeneous  mass  of  earth  deprived  of  cohesion  and 
unlimited  in  every  direction,  and  thus  determined  the  pres- 
sures erected  upon  the  bounding  surfaces  of  the  element. 
From  this,  when  a  plane  surface  limiting  a  mass  of  earth  was 
given,  the  direction,  magnitude,  and  position  of  the  total 
pressure  upon  this  plane  surface  could  be  easily  determined. 
Such  an  analytical  method  was  followed  by  Moseley,  Wink- 
ler,  Levy;  Considere,  and  Weyrauch.  However,  it  has  re- 
mained largely  in  the  field  of  scientific  investigations  and 


30        EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 

has  not  been  so  extensively  used  for  practical  purposes  as 
the  theory  of  the  sliding  prism. 

In  this  and  the  following  sections  is  given  a  method  for 
determining  earth  pressure  graphically,  according  to  the 
method  of  Professor  Rebhann,  based  upon  the  principle  of 
the  sliding  prism. 

GRAPHICAL  METHOD   AFTER   REBHANN 

12.  Let  a  bank  of  earth  held  up  by  a  retaining  wall  be 
represented  by  its  profile  ABDC,  Fig.  12,  and  let  the  follow- 
ing assumption  be  made : 

1.  The  pressure  against  the  wall  is  caused  by  a  prism 
of  earth  which  tends  to  separate  from  the  bank  and  slide 
along  a  plane  surface.     Such  an  assumption  is  not  strictly 
accurate,  because  the  surface  of  sliding  or  surface  of  rupture 
is  not  a  plane  but  a  cylindrical  surface,  having  a  cycloid  as 
directrix.     But  by  assuming  the  surface  of  sliding  to  be  a 
plane  surface,  we  greatly  simplify  calculations  and  the  results 
are  very  close  to  the  actual  facts,  erring  a  little  on  the  safe 
side  (since  the  sliding  prism  as  assumed  is  a  little  larger  than 
the  mass  which  actually  tends  to  split  off). 

2.  The  specific  weight   of   the   material   be  uniform   all 
through  the  mass  of  earth  forming  the  embankment.     This 
assumption  also  is  not  strictly  correct,  because  the  specific 
weight  varies  with  depth,  humidity,  etc. ;  but  the  laws  of  the 
variation  are  still  unknown  and  the  error  involved  in  consid- 
ering the  embankment  to  be  of  constant  specific  weight  is  so 
small  as  to  be  negligible  in  practice. 

3.  The  earth  is  devoid  of  cohesion.     Usually  the  earth 
behind  a  retaining  wall  is  filled  in  after  the  wall  is  built, 
so  that  in  most  cases  the  cohesive  force  has  been  entirely 
destroyed. 


RETAINING   WALLS 


31 


FIQ.  12. 


In  Fig.  12,  AB  is  the  back  of  a  retaining  wall  supporting 
a  mass  of  earth  limited  above  by  a  cylindrical  surface  whose 
profile  is  represented  by  an  irregular  line  as  shown. 

Since  it  is  assumed  that 
the  earth  is  entirely  devoid 
of  cohesion,  the  prism  of 
earth  above  the  plane  of 
natural  repose  would  have  a 
tendency  to  separate  from 
the  mass,  and  slide  along  the 
plane  of  repose.  But  in 
order  to  slide,  the  prism 
would  also  slide  along  the 
face  AB,  the  back  of  the  wall,  and  it  thereby  develops  an 
upwardly  directed  force  of  friction  which  will  counteract 
in  part  the  falling  tendency  of  the  prism.  In  consequence, 
instead  of  sliding  along  the  plane  of  natural  slope  it  would 
slide  along  the  plane  AD,  making  with  the  horizontal  AE 
an  angle  greater  than  the  angle  of  natural  repose  <£. 

Let  P  =  the  pressure  against  the  wall, 

$  =  the  angle  of  natural  repose  of  the  earth, 

and  <//  =  the  angle  of  friction  between  earth  and  the 

masonry  of  the  wall. 

Then  the  forces  acting  upon  the  prism  ABD  are :  the 
weight  W  of  the  prism ;  the  reaction  N  of  the  lower  mass  of 
earth,  normal  to  the  plane  of  sliding  ;  the  friction  F  produced 
by  the  normal  pressure  N\  the  reaction  N'  normal  to  AB, 
which  the  wall  opposes  to  the  prism ;  and  the  friction  F1 
between  wall  and  earth,  depending  upon  W. 

We  know  that 


32       EARTH   SLOPES,   RETAINING   WALLS,   AND  DAMS 

since  <f>  is  the  angle  of  friction  between  any  two  portions  of 

the  earth  ;  also 

F'  =  N'  tan  <£'. 

The  resultants  P  and  Q  of  the  pair  of  forces  F',  N1  ',  and 
F,  N,  are 

Q  = 

and 

P  = 


Substituting  for  F  and  F'  their  values,  we  have 


and 
But 


tan2  6  =  sec2  <f>  = 


cos2  (f> 
and 

1  +  tan2  <£'  =  sec2  <£'  =  —! 

COS2  (/>' 

so  that 


^cos24> 
and 


or 


cos  9  cos  9' 

This  means  that  the  prism  ABD  is  in  equilibrium  when  the 
angles  made  by  the  resultants  Q  and  P  with  the  normals  are 
respectively  equal  to  </>  and  <£',  and  intersect  the  line  of  action 
of  the  weight  of  the  prism  in  the  same  point  0. 

We  may  deduce  the  general  rule  that  the  pressure  P  of 
the  earth  against  the  back  of  the  wall  is  never  normal  to  the 


RETAINING   WALLS  33 

back  of  the  wall,  but  has  a  downward  slope,  making  with  the 
normal  an  angle  <f>f  equal  to  the  angle  of  friction  between 
the  earth  and  the  wall. 

13.  Values  of  <|>  and  $'.  — It  is  very  easy  to  find  by  experi- 
ment the  value  of  <£  for  different  soils,  but  it  is  not  so  easy 
to  find  the  value  of  <£'  because  we  have  no  satisfactory  method 
of  measuring  it,  and  because  it  varies  greatly  with  the  differ- 
ent kinds  of  earth  and  masonry. 

Colonel  Ande  assumes  <ft  =  26°  34' ;  Poncelet  and  Moseley 
have  given  to  <£'  the  values  of  27°  2',  18°  47',  and  21°  48' ; 
Scheffer,  Rebhann,  Otto,  and  Curioni  assume  </>'  =  <f> ;  while 
still  others  make  <j>f  =  ^  <£.  The  older  authors,  as  Coulomb, 
Prony,  and  others,  did  not  consider  $'  at  all,  thus  making  it 
equal  to  zero.  Following  Professor  Rebhann  we  shall  assume 
<f)f  =  <f>.  The  surface  of  the  back  of  the  wall  never  is  smooth, 
but  rather  is  very  rough.  When  the  earth  backing  is  rammed 
against  this  rough  surface,  it  fills  all  the  cavities  and  recesses 
in  the  masonry.  If  sliding  took  place,  evidently  the  earth 
could  not  disengage  itself  from  the  irregularities  in  the 
masonry,  but  rather  a  layer  of  earth  just  along  the  masonry 
would  stay  in  place,  and  the  prism  would  slide  along  this 
layer  of  earth.  This  condition  would  be  a  sliding  of  earth 
on  earth,  for  which  the  friction  factor  is  <f>.  It  seems  proper, 
therefore,  to  assume  that  the  friction  between  earth  and  back 
of  wall  be  the  coefficient  <£'  =  <f>. 

14.  Location  of  Plane  of  Rupture.  —  Consider  an  embank- 
ment in  equilibrium  or  just  before  the  caving  of  its  face; 
suppose  its  length  (in  Fig.  13  perpendicular  to  the  drawing) 
isl. 

Let  AD  represent  the  plane  of  rupture.  We  will  denote 
by  yS  the  angle  that  the  plane  of  rupture  makes  with  the 


34       EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 


FIG.  13. 


natural  slope.  The  prism  BAD,  sliding  along  the  planes 
AD  and  AB,  acts  as  a  wedge,  developing  the  pressures  P  and 
Q,  whose  directions  depend  upon  the  corresponding  angles  </> 

and  <£',  and  whose 
intensities  depend 
upon  the  weight  W 
of  the  prism.  We 
shall  limit  our  in- 
vestigation to  the 
forces  W,  Q,  and  P. 
Their  directions  are 
first  to  be  found, 
for  when  these  are 
known,  the  numer- 
ical amount  of  Q  and  P  is  easily  calculated,  thus  : 

Draw  a  vertical  line  (Fig.  14)  of  length  ab  =  TF;  from  a 
draw  a  parallel  to  P,  and  from  b  a  parallel  to  Q.  Then  the 
triangle  abc  is  the  triangle  of  forces,  be  representing  the 
value  of  Q  and  ac  the  value  of  P. 

It  remains  to  fix  the  plane  of  rupture  AD,  in  order  to  de- 
termine the  directions  of  P  and  Q.  Suppose  the  plane  AD 
turns  around  A  by  a  very  small  angle  d/3 ;  repre- 
sent its  new  position  by  AD'  (d/3  is  exaggerated 
in  Fig.  14).  The  weight  W  of  the  prism  will 
be  diminished  by  a  very  small  quantity  dW, 
represented  by  bb'  in  the  triangle  of  forces  (Fig. 
14).  Professor  Rebhann,  to  whom  this  demon- 
stration is  due,  assumes  that  upon  slightly  varying 
the  plane  of  rupture  the  pressure  P  remains  con- 
stant and  only  Q  varies,  its  new  value  being  Qf;  Qf  is  repre- 
sented by  cb'  in  the  triangle  of  forces  (Fig.  14),  the  angle 
bcb'  being  equal  to  d/3. 


RETAINING   WALLS  35 

In  Fig.  13  the  angle  which  Q  makes  with  N  (the  normal 
to  the  plane  of  rupture)  is  equal  to  $,  the  angle  of  internal 
friction  of  the  soil,  equal  to  the  angle  of  repose.  Now  the 
angle  a  =  y  -f  <£,  being  an  exterior  angle  of  the  triangle. 
Also  a=c£  +  /3,  as  the  angles  a  and  the  angle  <f>  +  @  have 
their  sides  respectively  perpendiculars.  Hence  we  have 

7=/3. 

Turning  the  plane  of  rupture  around  .A  by  an  angle  d/3,  the  re- 
sultant Q  is  turned  through  a  corresponding  angle  dy,  and 
for  the  same  reason  as  above  we  will  have 

0  +  d£=7  +  dy, 
and  since  /3  =  7, 

therefore  d/3  =  dy. 

But  the  angles  dy  and  bob'  are  equal,  having  their  sides 
respectively  parallel;  in  consequence 

d/3=bcbr. 

In  the  construction  of  the  triangle  of  forces  we  may  use 
such  a  scale  that  Q  be  represented  by  the  line 

be  =  AD. 
In  this  case 

Abcb'  =  A  ADD', 

and  having  already  assumed  that 

W=abzndbV  =  dW, 
we  will  have 

area  ABD  :  area  ADD'  =  ab  :  bbr, 

because  the  weights  of  prisms  of  the  same  material  and  equal 
height  are  to  each  other  as  the  areas  of  their  bases. 


36        EARTH  SLOPES,   RETAINING  WALLS,   AND  DAMS 

Further,  from  the  triangle  of  forces  we  have 

area  abc  :  area  bob'  =  ab  :  bbf. 

Combining  the  two  proportions  gives  the  proportion 
area  BD  :  area  AADDf  =  area  abc  :  area  bob'. 
Since  we  have  drawn  the  triangle  of  forces  to  such  a  scale 

area  ADD'  =  area  bcbf, 
it  will  result  that 

area  BAD  =  area  abc. 

Therefore  Q  makes  with  the  normal  to  AD  an  angle  <f>  and 
with  the  vertical  W  an  angle  D A  0  —  </>  =  ft.  ^ 

If  we  apply  the  triangle  of  forces  to  the  cross  section  in 
Fig.  13,  in  such  a  position  that  b  coincides  with  J.,  and  if  we 
then  turn  it  through  an  angle  90  —  c/>,  the  line  be  =  AD  will 
coincide  with  AD  and  ba  with  AK\  the  line  ca,  which  in  the 
original  position  made  an  angle  90  —  <£'  with  the  back  of  the 
wall,  after  rotation  will  make  an  angle 

90°  -  <£  +  90°  -  <t>f  =  180°  -  (<£  +  <£'), 

and  consequently  the  line  ac  will  fall  upon  DK.  Its  direction 
is  parallel  to  the  line  AH  drawn  from  A  at  an  angle  (f>  -f-  <f>f 
with  the  back  of  the  wall  AS. 

The  triangle  ADK  will  then  be  equal  to  the  triangle  abc. 
Since  we  already  know  that 

area  ABD  =  area  abc, 
it  follows  that 

area  ABD  =  area  ADK. 

That  is,  the  plane  of  rupture  AD_&sides  the  surface  ABDK 
into  two  equal  areas. 

The  line  AH  parallel  to  DK-aud  at  angle  </>  +  <#>'  with  the 
back  of  the  wall  is  called  the  directrix. 


RETAINING  WALLS 


37 


15.  The  Triangle  of  Pressure.  —  From  the  triangle  of  forces 
we  have 

W:QiP  =  alilciac  =  AK:  AD  :  DK.  (1) 

From  these  proportions  we  may  reduce  the  following 
principle : 

The  weight  of  the  sliding  prism,  the  reaction  of  the  plane 
of  rupture,  and  the  resistance  opposed  by  the  back  of  the 
wall  are  proportionals  to  the  three  sides  of  the  triangle  ADK. 

In  Fig.  15,  with  center  in  A  and  radius  AD,  draw  an  arc 


FIG.  15. 


DD',  so  that  AD'  =  AD  ;  and  with  center  at  K  mark  off 
KI=  DK.     Draw  DI,  and  DD"  then  will  have 

AK:  AD"  :  KI  =  A  ADK:  A  ADD"  :  ADKI, 

or,  which  is  the  same, 

AK:  AD  :  DK  =  A  ADK:  A  ADD"  :  A  DKL 
From  equation  (1)  we  have 

AK:AD:  DK  =  W:  Q  :  P; 
substituting  these  values  gives 

W:  Q  :  P  =  A  ADK-.  A  ADD"  :  A  DKI, 


38        EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 

and  as  the  area  of  AADK  =  A  ABD  is  proportional  to  the 
weight  W,  the  triangles  ADD"  and  DKI  will  be  respectively 
proportional  to  Q  and  P. 

The  reaction  Q  of  the  plane  of  rupture  is  equal  to  the 
weight  of  a  prism  having  ADD"  for  base  ;  or,  denoting  by  7 
the  weight  of  a  cubic  unit  of  the  soil, 

Q  =  AADD"  x  7. 

The  pressure  P  is  equal  to  the  weight  of  a  prism  of  unit 
height  having  the  triangle  DKI  as  its  bases, 

or 

P=ADKIx  7. 

The  triangle  DKI  is  called  the  triangle  of  pressure. 
The  demonstration  last  given  (Fig.  15)  is  due  to  Professor 
Weyrauch. 

APPLICATION  OF  THE   METHOD   TO   VARIOUS  PRACTICAL 

CASES 

16.  In  order  to  utilize  the  theory  just  given  in  solving 
practical  problems,  we  will  study  the  graphical  construction 
for  determining  both  the  plane  of  sliding  and  the  triangle 
of  pressure  in  various  embankments  : 

y  1.  When  the  embankment  is  surcharged,  its  upper  sur- 
face being  inclined  at  any  angle. 

2.  When   the    embankment   has  no  surcharge,  its  upper 
surface  being  horizontal. 

3.  When   the  surcharge   is   maximum,  its  upper  surface 
sloping  at  the  angle  of  repose  of  the  soil. 

4.  When    the    upper    surface    of    the    embankment   is   a 
broken  surface,  composed  of  several  planes  so  that  its  profile 
forms  a  broken  line. 


RETAINING   WALLS 


39 


s   a 


5.  When   the   upper   surface    of   the   embankment 
cylindrical  surface,  whose  profile  is  a  curve. 

The  angle  which  the  upper  surface  of  the  embankmentmakes 
with  the  horizontal  will  here  be  called  the  angle  of  surcharge. 

17.    CASE  I.  Surcharged  Embankment;  Any  Angle  of  Sur- 
charge. —  Let  ABC,   Fig.   16,   be  the  cross  section   of  the 


FIG.  16. 

bank  of  earth  limited  above  by  the  plane  BC,  AB  being  the 
back  of  the  retaining  wall. 

Supposing  the  problem  to  have  been  already  solved,  AD 
will  be  the  plane  limiting  the  prism  causing  the  pressure. 
From  A  draw  the  directrix  AH,  making  with  the  back  of 
the  wall  AB  an  angle  equal  to  <£  +  <£'.  From  D  draw  DK 
parallel  to  AH  \  from  ^Tdraw  KL  parallel  to  AD;  and  from 
B  draw  the  line  BM parallel  to  AH. 

The  two  triangles  ADIT  and  ADL  are  equivalent  because 
they  have  the  same  base  and  are  included  between  parallels. 
Also,  since  A  ABD  =  A  ADK  (because  the  plane  of  rupture 
divides  the  area  ABDKinto  two  equivalent  triangles),  the 
triangles  ABD  and  ADL  must  be  equivalent.  Considering 


40        EARTH  SLOPES,   RETAINING  WALLS,   AND  DAMS 

BD  and  DL  as  the  bases  of  these  triangles,  they  have  the 
same  altitude  (a  normal  from  A  to  BL),  and  hence  their 
bases  must  be  equal,  or  BD  =  DL. 

In  the  similar  triangles  ADO  and  CLK, 

AK:AC=DL  :  DC=  BD  :  DO. 
Also,  in  the  similar  triangles  BMC  and  DKC, 

BD:DC=MKiKC. 

Therefore, 


The  two  latter  quantities  may  be  written  in  terms  of  the 
former  as  follows  : 

MK=  AK-  AM  and  KO=  AC-  AK, 

which  substituted  in  the  previous  equation  give 
AK-.  A0=  AK-  AM:  AC-  AK, 
which  proportion,  multiplied  out,  gives 


This  means  that  AK  is  a  mean  proportional  between  the 
segments  AM  and  AC.  Therefore  by  drawing  a  semicircle 
on  A  C  as  diameter,  erecting  a  perpendicular  MN_&k~N,  and 
marking  off  AK=  chord  AN,  we  determine  the  point  K. 
For  the  triangle  ANC  is  a  right  triangle,  being  inscribed  in 
a  semicircle;  and  in  a  right  triangle  either  leg  is  a  mean 
proportional  between  the  adjacent  segment  of  the  hypote- 
nuse cut  off  by  a  perpendicular  from  the  vertex  to  the 
hypotenuse  and  the  whole  hypotenuse. 

^  This  gives  a  simple  means  for  locating  the  plane  of  rup- 
ture. AC  being  the  plane  of  repose,  draw  BM  parallel 
to  the  directrix  AH.  When  drawing  the  semicircle  ANC 


RETAINING  WALLS 


41 


locate  K  as  described ;  draw  KD  parallel  to  the  directrix 
AH.  Draw  AD  which  is  the  plane  of  rupture. 

To  describe  the  triangle  of  pressure :  with  center  at  K 
and  radius  KD,  lay  off  KI=  KD ;  unite  I  with  D.  The 
triangle  TCP T  will  Ji^thejfcrjanglejpf  pressure  or  the  trianglp. 
whose  area  multiplied-  by- -ihe-unit-  of  weigh%-ef-  the_material 
gives  in  pounds  the  total  pressure  against-.the-retaining  wall, 
per  lineal  foot  of  wall. 

Point  K  may  also  be  located  as  shown  in  Fig.  17 :  de- 
scribe a  semicircumfereiice  on  MO  as  diameter,  and  from  A 


FIG.  17. 

draw  the  tangent  AT.  Revolving  AT  about  A  into  AC 
gives  AK.  This  method  is  based  on  the  theorem 
of  geometry  that  the  tangent  to  a  circle  is  a  mean 
proportional  between  the  whole  secant  and  its  internal 
segment. 

Similar  methods  of  construction  leading  to  exactly  the 
same  result  can  be  made  with  the  use  of  other  lines  as  base 
in  place  of  AC: 

(#)  The  upper  surface  of  the  embankment  BO. 

(6)  The  back  of  the  retaining  wall  AB. 
The  directrix  AH. 


42        EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 


(a)  In  Fig.  18  project  the  points  A,  M,  and   0  upon  BC 

or  BO  produced  by  lines  parallel  to   the  directrix,  which 

gives    the   points    H,   B,   and    (7.     Since  the   equation   of 
Fig.  16,  viz., 


is   true  also  for  any  set  of  projections  of  these  lines,  the 
construction  above  applied  to  AC  and  AM  may  be  applied 


6  ... 


\  £^'" 
X 


I          |     >'/       \  \ 

/    \!  \  ^ 


FIG.  18. 

directly   to   HQ  and   HB,    giving   HD    according    to    the 
equation, 

HD*  =  HBx  HO. 


This  gives  a  point  which  is  the  projection  of  K  by  a  line 
parallel  to  AH,  and  as  will  be  seen  by  referring  to  Fig.  15 
this  is  the  point  D  in  the  plane  of  rupture.  Therefore  the 
construction  of  Fig.  18  gives  point  D  and  defines  the  plane 
of  rupture  directly.  The  details  of  the  construction  are 
shown  in  Fig.  18. 


RETAINING  WALLS 


43 


Having  Z>,  draw  DK  parallel  to  AH  to  intersect  the  sur- 
face of  repose  AC  at  K.  Then  lay  off  KI =  KD.  Triangle 
DKI  is  the  triangle  of  pressure. 

In  Fig.  18,  for  convenience,  the  value  of  0  has  been  taken 
larger  than  in  former  cases  and  the  resulting  triangle  of 
pressure  is  smaller. 

The  construction  of  Fig.  17  is  applicable  here  also,  of 
course,  as  shown  in  Fig.  18.  Describe  a  semicircle  on  BO  as 
diameter,  draw  a  tangent  from  H,  and  revolve  HE  to  HD. 


FIG.  19. 

(5)  Project  the  points  A,  M,  and  <7,  Fig.  19,  on  the  back 
of  the  wall  AB  by  lines  'parallel  to  BO,  giving  the  points 
A,  M',  and  B.  As  in  the  preceding  case,  the  construction 
for  a  mean  proportional  is  applied  to  AB  and  AM1,  giving 
the  point  K',  which  projected  by  a  line  parallel  to  BO  gives 
the  point  K. 

The  construction  is  clearly  indicated  in  Fig.  19.  Point 
M  is  found  as  before  by  projecting  B  upon  the  surface 
of  repose  AC  by  a  line  BM  parallel  to  the  directrix. 


44       EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 

(<?)  Referring  to  Fig.  20,  project  the  points  H,  B,  and  C 
on  the  directrix  AH  by  lines  parallel  to  AC,  which  give 
points  H,  B1,  and  A.  Then  HB'  and  HA  are  the  segments 
for  which  a  mean  proportional  is  to  be  found.  By  using  the 
construction  already  fully  explained,  and  shown  for  this  case 
in  Fig.  20,  point  Df  is  found,  which  is  projected  parallel  to 
A  C  upon  line  BO,  giving  point  D  in  the  plane  of  rupture. 
From  D  the  triangle  of  pressure  is  found  as  before. 

c 


FIG.  20. 

The  last  two  constructions,  (&)  and  (<?),  are  very  con- 
venient when  the  angle  of  surcharge  is  large,  as  that  point  0 
falls  outside  the  sheet  of  paper. 

18.  CASE  2.  No  Surcharge.  —  Professor  Weyrauch  has 
demonstrated  analytically  that  when  the  upper  surface  of 
the  embankment  is  horizontal,  i.e.  when  there  is  no  sur- 
charge and  the  back  of  the  wall  is  vertical,  the  pressure  of 
the  earth  against  the  wall  is  normal  to  the  back  of  the  wall; 
in  other  words  </>'  =  0. 

In  this  case  the  plane  of  rupture  and  the  triangle  of 
pressure  can  be  determined  very  simply  as  indicated  in 
Fig.  21.  The  directrix  AH  now  makes  the  angle  <f>  with 
the  back  of  the  wall;  then  the  two  triangles  AHB  and  AHO 


RETAINING  WALLS 


45 


will  be  similar  because  they  have  the  angle  at  H  in  common, 
and  the  angle  HO  A  =  BAH  because  both  are  equal  to  <£; 
the  third  angle  in  each  triangle  is  a  right  angle. 
The  triangles  being  similar, 

HB:HA  =  HA  :  HO, 
or 

HA*  =  HBx  HO. 


But  we  know  also  (from  Fig.  17)  that 


FIG.  21. 

It  follows  from  these  two  equations  that  HA  =  HD  and 
consequently  the  angles  HAD  and  HDA  are  equal.  Now, 
since  HO  and  AE  are  parallel,  angle  HDA  equals  angle 
DAE  as  they  are  alternate  interior  angles;  hence  we  may 
write  the  equations: 

Z  HAD  =  Z  DAE, 
or, 

/-HAB  +  ZBAD  =  Z.DAO+  Z  OAE, 

or,  what  is  same, 

<t>  +  BAD  =  DAO '  +  fa 
whence,  finally, 


46        EARTH   SLOPES,   RETAINING  WALLS,   AND   DAMS 


That  is  to  say,  the  plane  of  rupture  AD  bisects  the  angle 
between  the  back  of  the  wall  and  the  natural  slope  of  the 
earth. 

When  point  D  has  been  found,  the  procedure  for  drawing 
the  triangle  of  pressure  is  the  same  as  already  described. 
From  D  draw  DK parallel  to  the  directrix  ATI  to  intersect 
the  plane  of  repose  at  K.  From  -fiTlay  off  on  KA  a  segment 
KI=  KD,  and  draw  ID.  Then  triangle  DKI  is  the  tri- 
angle of  pressure. 

19.  CASE  3.     Embankment    with    Maximum    Surcharge; 
Angle  of  Surcharge  equal  to  Angle  of  Repose.  —  When  the 
angle  of  surcharge  is  equal  to  the  angle  of  repose,  i.e.  the 
top  of  the  embankment  is  parallel  to  the  natural  slope  of 

the  soil,  the  lines  BO  and 
AM  (Fig.  22)  are  parallel, 
which  means  that  point  (7, 
their  intersection,  is  at  in- 
finity. Under  these  condi- 
tions the  plane  of  rupture 
will  coincide  with  the  plane 
of  natural  slope  AM. 

The  triangle  of  pressure 
DIK,  being  contained  be- 
tween the  two  parallels  BO 
and  AC,  can  be  constructed 
at  any  point.  Draw  the  directrix  AH\  from  B  draw 
BM  parallel  to  AH\  make  M 0  =  MB  and  unite  0  with  B. 
The  triangle  BMO  is  the  triangle  of  pressure. 

20.  CASE  4.    Embankment  with  Irregular  Surcharge  (Top 
of  Embankment  a  Polygonal  Profile) .— Suppose   the   earth 
be  limited  above  by  two  planes  whose  traces  BCr  and   G-O, 


FIG.  22. 


RETAINING   WALLS  47 

Fig.  23,  form  a  broken  line.  It  is  necessary  to  change  the 
broken  profile  into  a  straight  line  by  the  method  of  trans- 
formation of  figures  into  simpler  equivalent  ones  as  given 
by  geometry.  This  reduces  the  problem  to  one  of  those 
already  discussed  (Cases  1  to  3). 

In  our  Fig.  23  produce  GrG  to  the  left,  connect  A  with 
6r,  and  from  B  draw  BF  parallel  to  AGr.  The  triangle 
AO-F  is  equivalent  to  the  triangle  ABCr  and  can  be  taken 
in  its  stead.  The  problem  is  thus  reduced  to  the  problem 


A 

FIG.  23. 

of  Case  1,  and  is  solved  as  follows:  from  F  draw  FM 
parallel  to  AH,  giving  point  M  on  the  surface  of  repose 
AC.  With  MO  as  diameter  draw  a  semicircle,  lay  a  tan- 
gent to  it  from  A,  AT,  locate  point  K  on  AC  by  making 
AK—AT,  and  from  K  draw  KD  parallel  to  AH,  giving 
point  D  on  the  upper  surface  FC.  Lay  off  the  segment 
KI=  KD  on  A  C,  and  connect  /  with  D,  which  gives  the 
triangle  of  pressure  KDI. 

When  the  upper  line  of  the  embankment,  instead  of  being 
a  broken  line  as  simple  as  the  one  just  considered,  consists 
of  many  segments,  it  still  may,  in  every  case,  be  reduced 
to  an  equivalent  straight  line  by  the  same  methods  as 
indicated  above. 


48        EARTH   SLOPES,   RETAINING  WALLS,   AND   DAMS 

21.  CASE  5.  Embankment  with  Irregular  Surcharge  (Top 
of  Embankment  of  Curvilinear  Profile).  —  When  the  embank- 
ment is  bounded  above  by  a  cylindrical  surface  whose  trace 
in  the  plane  of  section  is  a  curve,  the  problem  cannot  be 
solved  directly,  but  may  be  solved  by  trial. 

According  to  Professor  Rebhann's  demonstration  the  prob- 
lem can  be  reduced  to  a  relatively  simple  one,  viz.  to  find 
on  the  curved  profile  BO  a  point  M  such  that  the  following- 
equation  be  satisfied: 

area  ABM  =  area  AMK, 

the  line  MK  being  drawn  parallel  to  H.     Along  the  curve 
BC,  Fig.  24,  take  a  large  number  of  successive  points  i,  n,. 


FIG.  24. 


Ill,  •••  etc.,  so  close  together  that  the  sectors  ABi,  lAn,  ii.-4.iii,. 
•••  etc.,  can  be  considered  as  triangles  and  their  areas  easily 
calculated.  Assume,  then,  any  convenient  scale  of  reduc- 
tion b  for  these  areas  such  that  on  dividing  each  area  by 


RETAINING  WALLS  49 

b  we  obtain  a  number  and  may  therefore  represent  the 
areas  by  segments  of  straight  line.  On  AB,  or  in  our  case 
on  AB  produced,  lay  off  the  segments  A,  1 ;  1,  2 ;  2,  3, 
proportional  to  the  areas  of  the  sectors  BAi,  lA n,  iiAin, 
etc.,  so  that  the  line  A§,  for  instance,  when  multiplied 
by  b  will  represent  the  area  between  AB  and  Am. 

Now  with  center  at  A  lay  off  the  segment  A1  on  the 
line  Ai  produced,  A2  on  line  All  produced,  AB  on  the 
line  ^.ni  produced,  etc.  Draw  a  curve  through  the  points 
so  found.  The  radius  vector  of  this  curve  represents  to 
scale  the  area  of  the  prism  of  earth  between  the  wall  AB 
and  the  radius  vector  itself. 

Also,  from  A  draw  the  directrix  AH  making  the  angle 
<f)  +  <f>f  with  the  back  of  the  wall.  From  each  one  of  the 
points  I,  n,  in,  .  .  .  into  which  the  curvilinear  profile  has 
been  divided,  draw  lines  iKv  n^Q,  iii-fi"3,  etc.,  parallel 
to  the  directrix  AH;  the  triangles  iAKv  iiAK^  iuAKB, 
etc.,  are  thus  obtained.  Measure  the  area  of  each  one  of 
these  triangles,  reduce  these  areas  on  the  same  scale  6, 
and  lay  off  the  resulting  values  on  the  corresponding  radii 
downward  from  A.  Join  the  points  so  obtained  by  a 
smooth  curve.  Now  each  radius  vector  divides  the  em- 
bankment into  two  parts,  a  sector  and  a  triangle ;  for  exam- 
ple, the  radius  Am  gives  the  sector  .RAm  and  the 
triangle  m^.^.  These  different  areas  corresponding 
to  each  radius  are  represented  by  the  two  curves.  Where 
the  curves  intersect,  it  is  evident  the  area  of  the  sector 
is  equal  to  the  area  of  the  triangle.  But  according  to  the 
theory  of  Professor  Rebhann,  this  is  the  condition  which 
defines  the  plane  of  rupture.  Consequently  the  point  M 
(corresponding  to  point  D  of  the  preceding  cases)  is  found 
by  uniting  the  point  of  intersection  of  the  two  curves  of 


,50     EARTH   SLOPES,   RETAINING   WALLS,   AND  DAMS 

area  with  A,  and  producing  it  till  it  strikes  the  curvilin- 
ear profile  of  the  top  of  the  embankment  at  M. 

The  triangle  of  pressure  is  found  as  before.  From  M  draw 
MK  parallel  to  the  directrix  AH,  lay  off  on  the  plane  of 
natural  slope  a  segment  KP  —  MK,  join  M  and  P,  and  the 
resulting  triangle  MPK  is  the  triangle  of  pressure.  The 
area  of  this  triangle,  multiplied  by  the  weight  of  a  cubic 
foot  of  the  material,  gives  the  thrust  of  the  earth  against 
the  retaining  wall,  per  lineal  foot  of  wall. 

Of  course  this  method  is  slightly  inaccurate,  since  the 
sectors  BAi,  iJ.il,  etc.,  were  considered  as  triangles,  that 
is,  the  short  arcs,  Bi,  in,  etc.,  were  taken  as  straight  lines. 
But  by  making  the  segments  Bi,  in,  etc.,  sufficiently  short, 
the  error  may  be  reduced  to  any  desired  degree. 

VARIATION   OF   PRESSURE   WITH    HEIGHT    OF   WALL  ; 
INTENSITY  OF  PRESSURE;  CENTER  OF  PRESSURE 

22.  In  the  following  section  we  will  investigate  the  man- 
ner of  variation  of  the  pressure  against  a  retaining  wall. 

Suppose  the  point  A,  the  foot  of  the  wall,  is  moved  up- 
ward along  AB,  i.e.  the  wall  is  decreased  in  height.  For 
the  new  height  of  wall,  both  the  plane  of  sliding  and  the 
triangle  of  pressure  will  be  altered  to  determine  these  quan- 
tities under  the  new  conditions,  the  construction  previously 
used  is  to  be  repeated.  Since  all  the  angles  remain  the 
same,  and  the  respective  lines  remain  parallel  to  their 
original  positions,  the  resultant  figures  will  be  similar. 
Therefore,  also,  the  base  as  well  as  the  altitude  of  the  tri- 
angle of  pressure  will  be  proportional  to  the  height  of  the 
wall  AB,  and  in  consequence  the  value  of  the  pressure  P 
must  be  proportional  to  the  square  of  the  heights  of  wall. 


RETAINING  WALLS 


51 


In  Fig.  25,  for  example,  if  A  is  the  middle  point  of  AB, 
from  A  draw  A 0'  and  AD'  parallel  to  AO  and  AD  re- 
spectively; the  triangles  ABO  and  A'B'O'  are  similar, 
also  the  triangles  ABD'  and  ABD,  having  their  sides 
respectively  parallel.  But  in  similar  triangles  the  homolo- 
gous sides  are  proportionals,  and  since  AB  =  |  AB  by  con- 
struction, the  other  sides  of  the  triangles  are  in  the  same 
proportion,  i.e.  BD'  —  \BD  and  BO'  =  \BQ.  In  the  simi- 
lar triangles  AD'  0'  and  ADC,  the  altitudes  D'M'  and  DM 
are  in  the  same  pro- 
portion as  the  sides, 


Then,  the  isosceles 
triangles  D' KT  and 
DKl,  having  two 
sides  respectively 
parallel  and  the  in- 
cluded angle  equal, 
are  similar,  and  since 

other   corresponding 

sides  will  be  in  the  same  proportion,  or  KT  =  J  KI.  Now, 
the  area  of  the  triangle  of  pressure  K' DT  corresponding  to 
the  height  of  a  wall  A  B  =  J  AB  is  given  by  the  formula, 
area  =  \ >D'M'  x  KT.  Since  D'M'  =  \  DM  and  KT  = 

we    have,    area 


FIG.  25. 


2222  4 
In  other  words,  the  area  of  the  triangle  D' KT  is  one  fourth 
the  area  of  the  triangle  DKL  But  the  area  of  the  triangle 
DKl  multiplied  by  7,  the  unit  weight  of  the  material,  gives 
the  total  pressure  against  the  wall  AB,  while  the  correspond- 
ing pressure  for  a  wall  of  the  height  A  B  =  |  AB  will  be  one 


52      EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 

fourth  the  area  of  the  same  triangle  multiplied  by  7,  or  the 
pressure  is  one  fourth  as  great.  Similarly,  for  a  wall  one  third 
the  original  height,  the  pressure  is  one  ninth,  for  a  wall  of 
double  height  the  pressure  is  four  times  as  great,  etc.  Thus 
we  see  that  the  pressure  of  earth  against  a  retaining  wall  is 
proportional  to  the  square  of  the  height  of  the  wall. 

23.  Intensity  of  Pressure.  —  We  may  also  derive  the  prin- 
ciple that  the  intensity  of  pressure  upon  any  element  of  the 
back  of  the  wall  (represented  by  the  line  AB)  is  proportional 
to  the  distance  of  the  element  below  the  top  of  the  wall. 

Let  y  be  the  distance  of  any  point  on  the  back  of  the  wall 
AB  below  the  top  B.  The  total  pressure  above  point  y  being 
proportional  to  the  square  of  the  height  y  can  be  repre- 
sented by  cy*,  where  c  is  a  constant  coefficient.  Now,  if  y  is 
increased  by  an  infinitesimally  small  quantity  dy,  the  total 
pressure  increases  correspondingly  by  the  quantity  2  cydy ; 
making  dy  =  1,  the  total  pressure  increases  by  the  amount 
2  cy,  which  is  directly  proportional  to  the  height  y.  Con- 
sequently the  pressure  per  unit  of  height  between  y  and 
y  -f-  dy  is  given  by  2  cy,  which  is  proportional  to  the  height  y. 

The  intensity  of  pressure  at  the  middle  point  of  the  back 
of  the  wall  is  obtained  by  dividing  the  total  pressure  against 
the  wall  by  the  height  of  the  wall.  For  if  half  the  height 
of  the  wall  be  called  y,  the  total  pressure  above  the  middle 
point  is  P  =  cy*,  and  the  intensity  of  pressure  at  the  middle 
point,  as  just  found,  is  pl  —  2  cy,  or 

a».a«^*«4&  a) 

Also,  the  total  pressure  upon  the  back  of  the  wall  for  its 
full  height  is  P  —  cy1  2, 

but  in  this  case  y1  =  AB, 

so  that  P=cAB2. 


RETAINING  WALLS  53 

The  average  pressure  on  AB  is  obtained  by  dividing  this 
equation  by  AB,  which  gives 


Comparing  this  with  equation  (1)  we  see  that  p  =pr,  or  the 
intensity  of  pressure  at  midheight  is  exactly  equal  to  the 
average  pressure  against  the  entire  wall  AB. 

The  value  of  the  intensity  of  pressure  at  the  lowest  point 
of  the  wall,  at  A,  is  double  the  intensity  of  pressure  at  the 
middle  point  A'  .  For,  as  above  shown,  the  intensity  of  pres- 
sure at  any  point  is  expressed  by  the  general  formula  2  cy. 
At  the  point  A,  therefore,  the  intensity  of  pressure  is  2  cAB, 
while  at  A'  (Fig.  25)  it  is  CAB;  the  former  is  twice  the 
latter.  But  also,  since  the  intensity  of  pressure  at  midheight 
equals  the  mean  intensity  of  pressure  on  the  entire  wall,  as 
above  shown,  we  see  that  the  intensity  of  pressure  at  the  base 
of  the  retaining  wall  is  twice  the  average  intensity  of  pres- 
sure against  the  wall. 

The  total  pressure  P  is  given  by  the  area  of  the  triangle 
of  pressure,  DKI,  multiplied  by  the  weight  of  a  cubic  foot  of 
the  material,  7.  If  we  call  a  the  altitude  of  the  triangle,  IK 
being  its  base,  the  total  pressure  P  therefore  is 


Using  this  value  of  P  to  express  the  average  intensity  of 
pressure  p,  and  the  intensity  of  pressure  at  the  base  p1,  we 


These  values  of  the  intensity  of  pressure  at  the  middle  and 
the  base  of  the  wall  can  be  represented  graphically.  In  Fig. 
26,  from  A  draw  a  line  perpendicular  to  AB  and  on  it  lay  off 


54       EARTH   SLOPES,   RETAINING    WALLS,   AND   DAMS 


a  segment  AZ  =  IK.  Along  AB  lay  off  another  segment 
Alj  —  a.  Connect  L  with  Z\  the  triangle  ALZ  is  equivalent 
to  the  triangle  of  pressure  DKI.  Now  connect  B  with  Z  and 
from  L  draw  L  V  parallel  to  BZ. 

The  triangles  ABZ  and  ALV  SXQ  similar,  consequently 
AV  =AL 

AZ     AB 

C 
D 


T  ?"•*?-.: 


z 

FIG.  26. 

But  by  construction  AL  =  a  and  AZ  =  IK,  so  that 

AV       a 


IK 


AB' 
a  x  IK 


from  which  we  find  that 


We  found  above  that  the  intensity  of  pressure  at  the  base  is 
p'  =        .  p — ;   therefore  jt/  =  7J.F",  or  in  other  words  the 

intensity  of  pressure  at  the  base  of  the  wall  AB  is  given  by  the 
segment  A  V  multiplied  by  7,  the  specific  weight  of  the  soil. 
The  intensity  of  pressure  at  the  middle  point  of  AB  is 
p  =  ±pf.  This  will  be  represented  by  one  half  of  the  segment 
A  V  multiplied  by  the  weight  of  a  cubic  foot  of  soil : 

=  1    ,^IVa.IK  =l    Av=    AV 
P      2  ^  '    2     AB         2  7  72* 


RETAINING  WALLS  55 

24.  Center  of  Pressure.  —  In  the  right  triangle  AB  V  the 
width  of  the  triangle  at  any  point  represents  the  intensity  of 
pressure  at  that  point;  that  is,  ordinates  drawn  from  AB 
to  the  line  1?  J^  represent  graphically  the  different  values  of 
p  —  ^cy.     The  total  pressure  upon  the  back  of  the  wall  will 
be  given  by  the  sum  of  the  pressures  on  the  successive  unit 
areas,  that  is,  it  is  equal  to  the  area  of  the  triangle  BVA.     It 
is  applied  at  the  center  of  gravity  of  the  triangle  ABV,  at 
§  AB  from  the  vertex  B,  and  at  \  AB  from  A.     Hence  we 
may  say  in  general : 

The  point  of  application  of  the  total  pressure  on  the  back 
of  a  retaining  wall  is  at  one  third  the  height  from  the  base. 

The  triangles  AB  J^and  DKI  are  equivalents,  for  A  DKI= 
ALAZ  because  constructed  with  equal  bases  and  altitudes. 

ALAZ  =  ALAV+  ALVZ 
and  as  ALVZ=&LVB, 

(having  equal  bases  and  altitudes,  their  vertices  being  on 
lines  parallel),  it  follows  that 

A  LAZ  =  LAB  +  LVB  =  BA  V. 
Hence  ADKI=  LAZ=  BAY. 

THE  EARTH  PRESSURE  REPRESENTED  BY  A  LINE 

25.  The  pressure  against  a  retaining  wall  can  be  repre- 
sented by  the  length  of  a  line  drawn  to  a  determined  scale. 

Let  h<ym  be  the  scale  of  reduction,  in  which  h  may  be 
taken  as  equal  to  1,  while  ym  is  the  unit  of  weight  of  the 
material  used  in  the  construction  of  the  retaining  wall. 
Also,  let 

p  =  the  length  of  the  line  representing  the  pressure, 
a  =  the  altitude  of  the  triangle  of  pressure  DKI, 


56       EARTH   SLOPES,   RETAINING   WALLS,   AND    DAMS 

ry  =  weight  of  the  unit  of  volume  of  earth  in  the  embank- 
ment.    Then  we  can  write 


P 


_%» 
7 


In  Fig.   27,  a  is   the    angle   that  DD!  makes  with  AH  or 
its  parallel  DK\  then 


cos  « 

Substituting   this  value  of  IK  in  the  former  formula,  we 
get 

P  =  ~ 

2  h-^cos  a. 
7 

For  simplicity  call  the  denominator  2  h  —  cos  a  —  n,  then 


Now    the  angle    a  —  $  —  8,    equal   to  the    angle   that   the 
direction  of   pressure  makes  with  the  horizontal.     This  is 

evident  if  from  A  and 
D  we  draw  the  vertical 
lines  AT'  and  DT;  for 
the  angles  KDT  and 
HAT'  are  equals;  but 
HAT  =  <£  +  <£'  -S  and 
TDK=  TDD1  +  D'DK 
=  <£  +  D'DK,  since  the 
angle  2DD'  =  c/>  (be- 
cause the  sides  are  re- 
spectively perpendicular)  ;  hence  the  angle  D' DK=  a,  will 
be  equal  to  <f>f  —  B. 


RETAINING  WALLS  57 

The   following    graphical   construction   can   now   be   de- 
veloped :   on  the  horizontal  line,  Fig.  27,  lay  off  a  segment 

AP  =  2  A  —  •     Then  from  A  draw  a  line  AQ  making  with 

AP  an  angle  <*  =  <£'  —  £.     From  P  draw  PQ  perpendicular 

to  AQ.     Then 

A  Q  =  AP  cos  a, 

and  substituting  the  value  of  AP, 

AQ  —  2  h  —  cos  a  =  n. 

Along  the  perpendicular  line  QP  produced  lay  off  a  seg- 
ment QU=DD'=a.  Connect  U  with  A  and  from  the 
point  U  draw  the  line  UR  perpendicular  to  AU  meeting  AQ 
produced  at  the  point  R.  In  the  triangle  A  UR  it  is  easily 
seen  the  perpendicular  UQ  will  be  the  mean  proportional 
between  the  two  segments  AQ  and  QR,  that  is, 


QR-, 

but  UQ  =  a,  and  A  Q  =  n,  so  that 


Thus,  with  the  scale  of  reduction  selected  as  described,  the 
length  of  the  line  QR  represents  the  total  pressure  against 
the  wall  AB. 

It  is  observed  that  when  the  value  of  n  increases  the 
pressure  P  will  decrease.  Now,  n  increases  with  a  decrease 
of  a  =  <f>'  —  S.  Since  <fi  is  a  constant,  the  variation  of  P  will 
depend  exclusively  upon  the  value  S,  which  represents  the 
inclination  of  the  back  of  the  wall  from  the  vertical.  In 
other  words,  the  pressure  will  assume  different  values  accord- 
ing to  the  various  inclinations  given  to  the  back  of  the  wall. 


58       EARTH   SLOPES,   RETAINING   WALLS,  AND   DAMS 

The  pressure  is  smaller  when  the  wall  is  inclined  toward  the 
slope  of  the  embankment,  and  the  pressure  is  least  when 
<£'  =  6. 

When  the  inclination  of  the  wall  is  outward,  the  angle  S 
changes  sign,  so  that  we  will  have  «  =  <£'  +  B.  The  value  of 
n  will  then  decrease  with  increasing  values  of  8,  and  the 
value  of  the  pressure  P  will  increase  greatly. 

From  these  considerations  we  may  deduce  that  a  wall  is  in 
better  condition  to  resist  the  pressure  when  it  is  built  with 
its  back  inclined  toward  the  embankment.  In  practical 
work,  walls  are  usually  built  with  the  back  vertical,  so  that 
S  =  0.  But  walls  with  a  slope  of  -J-  or  -J^  will  be  more 
economical,  quite  as  convenient,  and  will  not  be  difficult  to 
construct. 

EFFECT  OF  COHESION  ON  PRESSURE  AGAINST  RETAINING 

WALLS  \m> 

26.  Hitherto,  the  problem  of  earth  pressure  against  re- 
taining walls  has  been  considered  without  regard  to  the 
cohesive  force  of  the  soil  or  backing  material.  This  was 
done  for  several  reasons,  chiefly  because  retaining  walls  are 
usually  built  to  support  filled  earth,  in  which  case  the  back- 
ing earth  is  practically  devoid  of  cohesion.  Even  in  the 
case  of  a  wall  supporting  a  bank  of  earth  in  its  natural  posi- 
tion, the  cohesion  is  usually  neglected,  because  the  value  of 
the  cohesion  is  easily  altered  by  atmospheric  influences, 
while  the  force  of  friction  remains  almost  constant  in  the 
same  embankment.  However,  the  forces  of  friction  do  not 
enter  into  action  until  the  cohesive  power  of  the  material 
has  been  entirely  destroyed  or  overcome. 

It  is  a  known  fact  that  a  bank  of  earth  endowed  with 
cohesion  exerts  a  smaller  pressure  against  a  retaining  wall 


RETAINING  WALLS 


59 


than  a  similar  bank  of  earth  without  cohesion.  Although 
for  simplicity  of  calculation  and  for  added  safety  the  cohe- 
sive force  of  the  material  is  always  neglected,  yet  there  are 
cases  in  which  it  will  be  useful  to  determine  how  far  the 
cohesion  of  the  earth  affects  the  total  pressure  against  the 
wall. 

Let  ABO,  Fig.  28,  be  a  bank  of  earth  in  which  AB  is  the 
back  of  the  retaining  wall,  AC  is  the  plane  of  natural  re- 
pose, and  AD  is  the  plane  of  rupture.  The  triangle  DKIis 

C 


FIG.  28. 

the  triangle  of  pressure  determined  according  to  the  methods 
indicated  in  the  preceding  sections. 

The  force  of  cohesion  will  prevent  the  prism  of  earth 
ABD,  which  causes  the  pressure,  from  sliding  along  the 
plane  AD.  Since  the  force  of  cohesion  is  uniformly  distrib- 
uted on  this  plane  of  sliding  and  is  proportional  to  the  area 
of  the  plane,  it  can  be  represented  by  the  plane  of  sliding 
itself,  and  in  our  case  will  be  represented  by  the  line  AD. 
In  all  these  calculations  the  depth  of  embankment  con- 
sidered, in  a  direction  perpendicular  to  the  plane  of  the 
drawing,  is  equal  to  1. 

To  determine  the  value  of  the  cohesive  force  of  the  mate- 
rial it  is  necessary  to  know  first  the  coefficient  of  cohesion. 


60       EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 

For  such  a  purpose  cut  a  trench  and  firid  by  experiment  the 
vertical  height  up  to  which  the  material  remains  in  equilib- 
rium without  support.  Let  AB',  in  Fig.  28,  be  such  a  ver- 
tical height;  then  from  B'  draw  B'O  perpendicular  to  the 
line  AC  of  natural  repose.  Lay  off  a  segment  AE  =  AB', 
and  from  E  draw  the  horizontal  line  EF.  We  know  that 
the  coefficient  of  cohesion  is 

K=\EF. 

The  force  of  cohesion  per  unit  of  surface  is  given  by 


and  the  total  force  of  cohesion  acting  to  prevent  the  prism 
which  causes  the  pressures  from  sliding  along  AD  will  be 
given  by 

AD. 


Now  resolve  the  force  of  cohesion,  represented  by  the  line  AD, 
into  its  two  components  Q  and  P,  as  was  done  for  the  prism 
ABD,  p.  31.  The  direction  of  Q  was  found  to  be  at  an  angle 
<£>  with  the  normal  to  AD,  hence  it  makes  with  AD  itself  an 
angle  90°  —  <£.  The  direction  of  the  pressure  P  was  found 
to  make  an  angle  <f>r  with  the  normal  to  the  back  of  the  wall, 
and  consequently  makes  an  angle  90°  —  <£'  with  the  back 
of  the  wall.  Therefore  draw  the  component  Q  to  make  an 
angle  90°  —  <j>  with  AD  and  the  component  P  to  make  an 
angle  90°  -  <£'  with  AB. 

Let  AL  and  DL  be  the  two  components  of  the  force  of 
cohesion  represented  by  AD. 

The  decrease  of  the  earth  pressure  as  a  result  of  the  cohe- 
sive force  of  the  material  will  be  represented  by  the  compo- 
nent AL,  that  is,  the  pressure  is  decreased  by  a  quantity 


RETAINING   WALLS  61 

proportional  to  AL  and  having  the  amount 

C 

which  can  be  written 


Such  a  quantity  can  be  graphically  represented  in  the  tri- 
angle of  pressure  as  follows  : 

Along  the  line  A  0  from  /  toward  IK  lay  off  a  segment 
IM  =  J  AL.  At  M  erect  a  perpendicular  to  A  C  and  make 
MN=  EF.  Connect  N  with  I.  Then  the  triangle  IMN 
represents  the  decrease  in  pressure  due  to  the  cohesion, 
because  the  area  of  this  triangle  multiplied  by  7  will  by  con- 
struction be  found  equal  to  (7r 

Transform  the  triangle  IMN  into  an  equivalent  triangle 
having  the  base  IK  common  with  the  triangle  of  pressure, 
and  the  opposite  vertex  resting  upon  one  of  its  sides.  Let 
the  triangle  IKN'  be  equivalent  to  IMN.  The  former  pres- 
sure represented  by  the  total  area  of  the  triangle  DKI  x  7  is 
reduced  through  the  effect  of  the  cohesion  by  a  quantity 


The  remaining  pressure  against  the  wall,  then,  is  repre- 
sented by 

P1  =  AIDNf  xy. 

If  instead  of  the  total  available  value  of  the  force  of  cohe- 
sion, only  a  portion  of  this  force  is  relied  upon,  the  amount 
of  the  pressure  decrease  will  be  determined  in  a  similar  man- 
ner, except  that  instead  of  the  whole  value  of  the  cohesion 
only  |  or  ^  should  be  taken,  depending  upon  the  factor  of 
safety  used.  Thus,  in  the  case  just  considered,  if  it  were  re- 
quired to  know  the  decrease  of  the  pressure  which  would 


62        EARTH   SLOPES,   RETAINING  WALLS,    AND   DAMS 

result  by  the  influence  of  half  the  available  force  of  cohesion, 
the  decrease  of  pressure  is  changed 

from  C  =    yEFxAL  to 


This  quantity  is  represented  graphically  on  the  triangle  of 
pressure  by  laying  off  from  I  toward  TTa  segment  equal  to 
\  AL,  erecting  a  perpendicular  and  making  MN=  EF  the  re- 
sulting triangle,  which  can  be  easily  converted  into  another 
having  IK  for  base,  and  the  vertex  along  DK  represents  the 
decrease  of  pressure. 

When  the  bank  of  earth  stands  in  equilibrium  on  the  slope 
AB  for  the  full  height  of  the  retaining  wall  to  be  constructed, 
the  force  of  cohesion  entirely  counteracts  the  pressure.  This 
means  that  the  triangle  IMN  is  equal  to  the  triangle  of  pres- 
sure IDK,  and  consequently  that  there  is  no  pressure  at  all 
on  the  back  of  the  wall.  In  this  case  the  wall  may  be  con- 
structed of  any  thickness  whatever,  without  regard  to  ques- 
tions of  earth  pressure. 

THE  PRESSURE  OF   PASSIVE  RESISTANCE  OF  THE  EARTH 

27.  In  the  whole  of  the  preceding  discussion  we  have 
considered  the  pressure  of  the  earth  against  a  retaining  wall 
on  the  assumption  that  the  wall  is  stable  and  immovable, 
and  that  the  pressure  is  caused  by  the  tendency  of  the  earth 
to  move  out  of  its  original  position.  But  it  may  happen  that 
the  wall  exerts  a  pressure  against  the  backing  earth  tending 
to  force  it  back  out  of  its  position.  In  this  case  the  earth 
exhibits  an  enormous  resistance,  far  greater  than  the  pres- 
sures heretofore  considered.  This  phenomenon  is  observed 
occasionally  in  the  abutments  of  bridges,  where  the  arches 


RETAINING   WALLS 


68 


tend  to  push  the  abutments  outward ;  if  the  thrust  of  the 
arches  is  greater  than  the  active  outward  pressure  of  the 
earth,  it  forces  the  abutments  back  and  develops  the  so-called 
passive  pressure  of  the  backing  earth. 

The  conditions  of  the  problem  are  the  same  as  those 
already  considered,  except  that  they  are  reversed. 

Let  a  wall  AB,  Fig.  29,  exert  a  pressure  against  the 
backing  earth  greater  than  the  active  outward  pressure  of 


the  earth.  Then  a  prism  ABE  will  tend  to  separate  from 
the  mass  and  slide  upward  along  the  plane  of  rupture.  In 
this  case  we  have  the  same  force  as  heretofore,  but  the 
weight  of  the  prism  ABE  will  now  oppose  the  ascending 
movement,  and  the  resistances  of  friction  along  the  sur- 
faces AB  and  AE  will  now  act  downward ;  consequently  the 
values  of  cf>  and  </>'  will  change  sign.  In  Fig.  29,  AB  being 
the  back  of  the  wall,  ABE  is  the  embankment  as  limited 
above  by  the  surface  BE,  and  AF  is  the  trace  of  the  plane  of 
natural  slope,  passing  through  A  but  sloping  downward  to 
the  right  because  <£  is  negative. 


64       EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 

Draw  the  directrix  AGr  making  with  AB  an  angle  equal 
to  <£+</>'  to  the  right,  since  <j>  and  <f>r  have  changed  signs, 
whereas  in  the  case  of  active  pressure  the  directrix  extended 
to  the  left  of  AB.  Produce  BE  until  it  intersects  the  plane 
of  natural  repose  AF  at  F ;  then  on  BF  as  diameter  describe 
a  semicircle.  From  G-  draw  the  line  G-T  tangent  to  this 
semicircle  and  make  CrE =  CrT.  Connect  E  with  A  ;  then 
the  line  AE  will  represent  the  plane  of  sliding  of  the  prism 
ABE. 

Having  located  the  plane  of  rupture  it  is  easy  to  deter- 
mine the  value  of  the  pressure.  From  E  draw  the  line  EK 
parallel  to  the  direction  AGr,  intersecting  the  line  of  natural 
repose  AF  at  K.  From  JK  lay  off  along  KF  a  segment 
KM  =  KE  and  connect  M  with  E.  The  triangle  KEM  will 
be  the  triangle  of  passive  pressure  or  the  triangle  whose  area 
multiplied  by  the  weight  of  a  cubic  foot  of  the  material  will 
give  in  pounds  the  value  of  the  resistance  of  the  bank  of 
earth  to  the  inward  pressure  of  the  wall. 

The  intensity  of  the  passive  pressure,  or  the  pressure  per 

square  foot  of  wall,  can 
be  found  in  the  same 
manner  as  for  the  active 
pressure,  as  follows  : 

In  Fig.  30,  along  the 
line  AZ  drawn  perpen- 
dicular to  the  back  of 
the  wall  AB,  lay  off  a 
segment  AZ  =  KM,  from 
FlG>  3°'  A  along  AB  produced 

lay  off  another  segment  AL  —  a,  the  altitude  of  the  tri- 
angle KEM.  Join  B  writh  Z  and  from  L  draw  L  V  parallel 
to  BZ.  Then 


RETAINING   WALLS  65 

AV-.AZ  =  AL  :AB, 
or,  substituting  for  AZ  and  AL  their  values, 


that  is, 

AVx  AB  =  a  x  KM, 

from  which 

a  x  KM 


AV 


AB 


Since  the  intensity  of  pressure  at  the  base  of  the  wall  p  is 
equal  to  twice  the  average  pressure,  or  twice  the  area  of  the 
triangle  of  pressure  multiplied  by  the  unit  of  weight  of  the 
material  and  divided  by  the  total  height  of  the  wall  AB,  we 
find 


Then  the  intensity  of  pressure  at  the  middle  of  the  height 
of  the  wall,  which  is  equal  to  the  average  pressure  along 
AB,  is 

,=  1      =l7ax  JOf=l    Ay 

P       2^      2      AB          27 

In  the  triangle  ABV,  the  ordinates  measured  perpendicu- 
lar to  AB  represent  the  intensities  of  pressure  at  the  various 
points  of  the  wall.  It  follows  that  the  center  of  pressure  is 
at  the  center  of  gravity  of  the  right  triangle  ABV,  that  is, 
.at  \  of  AB  from  A. 

In  order  to  compare  the  active  and  passive  pressures,  in 
Fig.  29,  draw  the  triangle  of  active  pressure  GrIH  in  the 
usual  manner.  This  is  found  to  be  much  smaller  than  the 
triangle  representing  the  passive  pressure.  The  difference 
increases  with  increase  of  the  angles  (f>  and  <//,  and  decreases 


66        EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 

with  decrease  of  the  same  angles ;  when  both  these  angles 
are  equal  to  zero,  as  in  the  case  of  liquids,  then  the  active 
and  passive  pressures  are  equal. 

The  difference  in  magnitude  of  the  active  and  passive 
pressures  depends  also  upon  the  inclination  and  direction  of 
the  back  of  the  wall  AB.  Thus,  for  instance,  in  walls  in- 
clined toward  the  outside,  the  difference  will  be  found  to 
be  smaller  than  in  walls  either  vertical  or  inclined  toward 
the  embankment. 

In  many  actual  cases  the  passive  pressure  of  earth  has 
been  observed  to  be  3,  4,  10,  even  20  times  greater  than  the 
active  pressure,  and  under  the  proper  conditions  it  may  be 
still  larger. 


CHAPTER   III 

RETAINING  WALLS    (continued):    ANALYTICAL   METHODS 

REBHANN'S  METHOD 

28.  BESIDES  the  demonstration  of  the  graphical  process 
for  calculating  earth  pressure  as  given  in  the  preceding 
chapter,  Professor  Rebhann  has  given  also  an  analytical 
demonstration  of  his  theory  in  the  following  manner  : 

Referring  to  Fig.  31,  let  AS  be  the  back  or  interior  face 
of  a  retaining  wall,  B O  the  limiting  surface  of  the  embank- 
ment, AD  any  possible  plane  of  rupture  passing  through  A, 
and  represent  by  W  the  weight  of  the  prism  ABD.  The 
prism  ABD  in  sliding  along  AD  will  exert  a  pressure  against 
the  face  AB  of  the  retaining  wall  and  against  the  plane  of 
sliding  AD.  Calling  P  and  Q  the  reactions  of  these  two 
planes,  for  equilibrium  the  three  forces  W,  P,  and  Q,  must 
meet  in  a  point.  Therefore,  resolve  the  forces  W,  P,  and  Q 
into  their  components  parallel  and  perpendicular  to  AD 
respectively.  Then  for  equilibrium  we  must  have  :  (1)  the 
algebraic  sum  of  the  components  parallel  to  AD,  as  well  as 
the  sum  of  those  perpendicular  to  AD,  must  be  equal  to 
zero ;  and  (2)  the  algebraic  sum  of  the  moments  of  the 
forces  with  respect  to  any  point  must  be  equal  to  zero. 

That  is, 

RZ  -  E2-  R  =0,  (1) 

N3  +  N2-N  =0,  (2) 

Pp-Qq=0,  (3) 

67 


68       EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 


where  w,  p,  and  q  are  the  distances   from  A  of  the  forces 
W,  P,  and  Q  respectively. 
From  (1)  and  (2)  we  have 


RS-R<,  =  R,  (4) 

jv8  +  jy2  =  jv;  (5) 

R  —  N  tan  (f>  =  (jV3  +  JVa)  tan  <£,  (6) 

Evidently  the  one  sliding  plane,  or  in  other  words  the  actual 
plane  of  rupture,  will  be  that  for  which  the  fraction 


but 
so  that 


has  the  greatest  value.     If  we  denote  by  a  the  angle  DWB' 
between  the  plane  AD  and  the  vertical  passing  through  TF, 

P 


FIG.  31. 


we  can  represent  Rz,  Ry  JV8,  and  N2  in  terms  of  a ;  and  if 
we  then  differentiate  the  above  fraction  in  respect  to  a,  and 
put  the  differential  equal  to  zero, 


da  \  N*  + 


(8) 


RETAINING  WALLS:     ANALYTICAL   METHODS         69 

we  will  obtain  the  condition  giving  the  maximum  value  of 
the  fraction,  and  therefore  the  condition  which  finds   the 
plane  of  rupture. 
To  do  this,  we  write 

J^=  TFcosa, 


-ZVg  =  TFsina, 
N%  =  P  cos  (a  -J-  (f>r  —  5). 
V  Substituting  these  values  in  (7), 

TFcos  a  —  P  sin  (a  -+-  </>'  —  S) 

from  which  we  deduce 


P  = 


TF  cos  a  —  W  sin  «  tan  <j> 


sin  (a  +  <£'  —  8)  -f-  cos  («  +  </>'  —  5)  tan  <£ 
TTr        cos  (<f>-h  «") 


or 

P  cos  (ft  +  «) 


>T     sin  <>  +  <£  +  </>'- 
and 


P  cos  (</>  +  «) 

Substituting  in  (8)  the  values  of  J23,  K2,  ^V3,  JV^,  we  have 

6?  f  TFcos  <*  -  P  sin  Q  +  <f>r  -  8)"|  ^  Q 
c?aL  Wsin  «  -  P  cos  («  +  <£'  -  S)J 

Differentiating, 

f^cos  «  -  Tfsin  a  -  P  cos  («  +  £'-  8)1 

Lrf«  J 

x  [  TTsin  «  +  P  cos  («+<#>'-  5)] 


70        EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 

-[TFcos<*-Psin(«  +  $  -  £)] 

X  [~—  sin  a+Wcosa-P  sin  (a  +  <£'  -  8)1=  0, 
L  d«  J 

which  can  be  written 

[  IF  sin  a  cos  a  +  P  cos  (a  +  $'  —  8)  cos  a  —  TTcos  a  sin  a  + 

P  sin  («  +  <£'-$)  x  sin  a]  ^  -  17  sin  a  +  P  cos  («  +  <f>'-S)2 

C^6C 

-Psin  («  +  <#>'  -3)2=0. 
This  can  be  reduced  as  follows  : 

P  cos  (#'  -  8)  d^-W2  +2WP  sin  (<£'  -  3)  -  P2  =  0. 
a« 

Dividing  by  TFP  and  changing  signs, 

*-  2  sin  (^'  -  g)  +  4-  cos  &  ~S)*     =  0.     (13) 
P  da 


But  (7  W=  J  Aday  or       -  =  1  JLZ>  7 

6tr^ 

which,  substituted  in  (13),  gives 


Now  substituting  for  —  and  -—their  values  as  found  in  (10) 

and  (11),  we  will  have 

sin    «  +  <>  +       -  S  x         x  cos        +  « 


cos  (<£  +  «)  sin  (a  +  0  +  <#>'- 

cos(4>'-S) 


. 
or 

sin  (c^>  4-  «)  cos  ((/>!  —  8)  +  cos  (  <fr  -f  <*)  sin  (d)1  — 
cu&    <>  +  «     • 


RETAINING   WALLS:    ANALYTICAL   METHODS         71 

_  2  sin  (  <f>'  —  £)  cos  (  <f>  +  ft)   ,  cos  (  <f>  +  a) 

cos  (<£  +  a)  sin  ((/>  +  a  +  <f>'  —  8) 


which  reduces  to 


sin  (<fr  +  ft  +  $'  —  &)  ,          cos  (  <ft  4-  «) 

cos  (<£  +  a)  sin  (<£  +  a  +  <£'  —  8) 


2W 

This  can  also  be  written  as  follows  : 

sin  ( <ft  +  a  —  <f>f  +  £)  sin  (<ft  +  ft  +  $  —  S)  +  cos2  f <ft  4-  «) 
cos  (c/>  -f  ft)  sin  (c/>  +  «  +  <//  —  8) 


or 

sin2  (ft  +  a)  cos2  (<£'  -  3)  —  cos2  (ft  +  ft)  sin2(<fr'  -8)4-  cos2(  <ft  +  ft) 
cos  (</>  +  a)  sin  (</>  +  ft+  </>'  —  8) 


which  may  be  reduced  to 

cos2    <'  -  B  cos      '  - 


cos  (</>  -j-  «)  sin  (<£  +  ce  +  <£'  -  S)  2  TT 

and  from  this  the  value  of  W  is  easily  deduced  as  : 
cos    *    +  a    s^n    <>  +  ft  +  <>r  —  8 


W 


cos2 


Substituting  this  value  in  (9)  the  value  of  P,  the  pressure  is 
found  to  be 


p  _  cos2  (<f>  +  ft) 


cos  (<£'  -8;         2 


72        EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 

The  forces  P,  W,  and  Q  being  in  equilibrium,  the  sura  of 
their  horizontal  components  must  be  equal  to  zero,  that  is, 

P  cos  (<£'  -  8)  =  Q  cos  (</>  +  a), 
or 

coS(^-8)  =  cos(  ^.  (15) 

cos  (<£  +  a)  2 

In  the  triangle  ADK,  Fig.  31,  from  D  draw  DD'  perpen- 
dicular to  AK  and  DJ!f  vertically.  Then  the  angle  ADD' 
will  be  equal  to  (<£  +  «);  for  angle  ADM=  B'AD,  their 
sides  being  respectively  perpendicular,  and  D'DM  is  equal 
to  <£  for  the  same  reason,  so  that 

D'DM=  ADM+  MDD'  =  <*  +  £. 

Further,  DM  is  parallel  to  AB',  while  DK  is  parallel  to 
AH-,    therefore  MDK=  HABr       But   HAB1  =  <t>  +  <£'  -  8  ; 
therefore  angle  MDK=  $  -f-  <£'  -  S.      Since   MDK=  MDD1 
+  D'DK=  <f>  4-  D'D^T,  it  follows  that  D'DK=  $  -  B. 
Now  in  the  triangle  AKD  we  have 

DD'  =A  D  cos  ((^  +  «)  (16) 

and 

(17) 


cos  (</>'  —  o) 
Consequently, 

cos     >  +  «    sin  (<fr  +  «  4-  $'  ~ 


__ 


COB  (*'-«)  2 

TF=  A  ABDy,  (18) 

which  means  that 

The  plane  of  rupture  AD  divides  the  surface  ABDK  into 
two  equivalent  parts. 
Draw  AF  perpendicular  to  DK\  then 

=  AD  sin 


RETAINING   WALLS:    ANALYTICAL   METHODS         73 

Dividing  equation  (16)  by  equation  (19)  we  get 

D  Dr  =         cos  (ft  +  «)         =  P 
AF      sin  (</>+«  +  <£'-£)      w' 

Since  by  construction  DK=  KM, 


AKDA       AF      sin  (<£  +  «  +  <£'  -  S)      Tf 

but  equation    (18)    shows   that   A  KDA<y  =  JF".      Therefore 
KDM<y  =  P,  or,  in  other  words  : 

The  pressure  of  the  earth  against  a  retaining  wall  is  equal 
to  the  weight  of  a  prism  having  the  triangle  KDI  as  a  base, 
or 


Finally,  A  ADD"  = 

and  since  by  construction  ADff  =  AD,  while  from  equation 
(16)  DD'  =  AD  cos  (<£  +  a),  we  have 


A  ADD"  =  -|  ZD  cos  (<£  +  «), 

which  is  the  value  obtained  for  Q  in  equation  (15). 
Therefore 


in  other  words,  the  reaction  Q  of  the  plane  of  rupture  is 
equal  to  the  weight  of  a  prism  of  earth  having  the  triangle 
ADD"  as  base. 

It  will  be  apparent  from  the  three  equations  W=  7  •  ABD 
=  7  •  ADK,  P  =  y  MDK,  and  Q  =  7  -  ADD",  that 

W'.Pi  Q  =  AK:DK:AD. 

We  can  therefore  express  the  relations  between  these  quan- 
tities by  saying  : 


74        EARTH   SLOPES,   RETAINING  WALLS,   AND   DAMS 

The  weight  of  the  sliding  prism,  the  reaction  of  the  retain- 
ing wall,  and  the  reaction  of  the  plane  of  rupture  are  in  the 
same  proportion  as  the  sides  of  the  triangle  AKD. 

FORMULAS  OF  RANKINE  AND  WEYRAUCH 

29.  Various  authors  have  given  different  theories  of  earth 
pressure  against  retaining  walls,  based  upon  entirely  dif- 
ferent principles.  Some  authorities  even  discard  all  theo- 
ries on  the  ground  that  they  involve  many  assumptions  and 
therefore  cannot  give  reliable  results.  It  is  true  that  there 
has  not  yet  been  devised  a  method  that  will  give  in  absolute 
and  unexceptionable  manner  the  value  of  the  pressure 
against  a  retaining  wall;  but  since  the  various  approved 
methods  lead  to  almost  the  same  result,  any  one  of  them  is 
capable  of  giving  a  result  within  certain  limits  satisfactory. 
In  any  actual  case  the  engineer  may,  by  comparing  the 
assumptions  made  in  the  course  of  the  calculation  with  the 
particular  conditions  of  the  bank  of  earth  under  considera- 
tion, secure  a  fair  judgment  of  the  pressure  that  the  wall 
to  be  designed  has  to  resist,  and  will  be  enabled  at  the  same 
time  to  select  a  suitable  factor  of  safety  for  the  design. 

After  the  pressure  of  the  earth  has  been  determined 
graphically  according  to  the  theory  of  Professor  Rebhann 
as  given  in  the  preceding  chapters,  it  will  be  instructive  to 
calculate  the  pressure  by  another  method  and  compare  the 
results. 

Professors  Rankine  and  Weyrauch  have  studied  the  problem 
of  earth  pressure  by  analytical  methods.  Their  theories 
require  a  long  and  complicated  discussion,  and  for  lack  of 
space  they  are  therefore  not  given  here.  The  resulting 
formulas  only  are  given. 


RETAINING   WALLS:   ANALYTICAL   METHODS          75 

Rankine's  Formulas.  —  Professor  Rankine  assumes  that 
the  inclination  of  the  pressure  is  always  parallel  to  the 
upper  line  of  the  profile  of  the  embankment,  or,  in  other 
words,  that  it  is  parallel  to  the  surcharge,  and  he  considers 
the  back  of  the  wall  vertical.  If  we  denote  by 

IF,  weight  of  the  earth  per  cubic  foot, 
A,  height  of  the  embankment  in  feet, 
«,  the  angle  of  surcharge, 

and  <£,  the  angle  of  repose  of  the  earth,  he  gives,  for  the 

general  case, 

n      Wh2  cos  a  —  Vcos2  a  —  cos2  <f> 

P  =  —  —  x  cos  a  —  —  -  ~. 

*  cos  «  +  Vcos2  a  —  cos2  </> 

For  level-top  embankments,  i.e.  where  a  =  0,  this  reduces  to 


•  •  Y;*  "1         1  -f  sin  $ 

For  surface  sloping  at  the  angle  of  repose,  i.e.  when  a  =  <£, 
it  becomes 


T>  . 

P  =—  —  cos  9. 

Weyrauch's  Formulas.  —  Professor  Weyrauch  has  dealt 
with  all  possible  cases  of  retaining  walls,  not  only  those  of 
different  surcharges,  but  also  those  of  the  inclined  wall,  the 
wall  leaning  either  forward  or  backward.  Using,  in  general, 
the  same  symbols  as  heretofore,  i.e. 

P=  total  pressure  of  the  earth  against  the  wall. 
<£  =  angle  of  repose  of  the  material. 

<f>f  =  angle  of  direction  of  pressure  with  the  normal  to  the 
back  of  the  wall. 

a  =  angle  of  the  surcharge  with  the  horizontal. 


76        EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 

S  =  angle  of  the  back  of  the  wall  with  the  vertical  taken 
either  positive  or  negative,  according  as  the  slope 
of  the  wall  will  fall  either  to  the  left  or  right  of  the 
vertical. 

h  =  height  of  the  wall. 
w  =  weight  of  earth  per  unit  of  volume. 
/5  =  angle  between  plane  of  rupture  and   the  vertical, 

and,  referring  to  Fig. 
B       ^*«"tr'    /  32,  he 


FIG.  32. 


p  = 


in  which 


-  S) 


General  Formulas. — 
Embankment  with  any 
surcharge. 

Wall  leaning  for- 
ward, i.e.  away  from 
the  embankment, 

wh 


2  cos  <'- 


n 


sn    <   - 


and  the  value  of  <f>'  is  deduced  from  the  equation 


tan  <f>'  = 


sn         ~  «  ~ 


-  cos  (2  S  -  a)  +  4  cos  2  (8  -  a)' 
in  which  the  value  of  A  is  given  by 

j  _  cos  a  —  Vcos2  a  —  cos2  0 

COS2</> 

Back  of  the  wall  vertical,  5=0, 


2  cos 


RETAINING  WALLS:   ANALYTICAL   METHODS          77 

Back    of    the    wall    inclined    toward    the   embankment,    B 
negative, 

p  _  /    cos  (<£>  +  8)    V  wfi? 

\(1  -f  n)  cos  —  Bj       2  cos  (<£'  4-  £)' 

the  values  of  n,  <£',  and  J.  being  deduced  from  the  formulas 
given  above,  except  that  B  is  negative. 
^Embankment  with  no  surcharge,  i.e.  «=  0, 

p  _        tan  B     _      wh2 

in  which  the  value  of  <f>r  is  given  by 
'      -  sin  2  * 


tan  <£'= 

1  -  sin  <£  cos  2  S 


When  the  back  of  the  wall  is  vertical,  8=0,  and  therefore 
by  the  formula  last  given,  <f>r  will  also  equal  zero  and  the 
value  of  the  pressure  reduces  to 


If  the  wall  is  required  to  retain  water  instead  of  earth,  so 
that  </>  is  equal  to  zero,  then  the  above  formula  reduces  to 


which  is  the  well-known  formula  for  the  pressure  of  water 
against  a  dam. 

The  pressure  cannot  be  calculated  when  the  wall  is  inclined 
inward. 

Embankment  with  surcharge  parallel  to  the  natural  slope; 

s-W  wit* 


cos 


8  2  cos 


78        EARTH   SLOPES,   RETAINING  WALLS,   AND   DAMS 

in  which  <//  is  obtained  from 

tan  A'  =     sin0cosC4,-28) 
1  -  sin  <f>  sin  (<£  -  2  S) 

When   the   back   of   the   wall   is    vertical,  3  =  0,  and   then 
(£  =  <£',  the  value  of  the  pressure  is  simplified  to  the  form, 


When  the  back  of  the  wall  is  inclined  inward  toward  the 
embankment,  the  value  of  S  becomes  negative  and  the 
general  formula  will  be 

wh* 


cos  -8  2  cos  (<£'-  8)' 

the  value  of  <£'  being  deduced  from 


1  -  sin  £  sin  (<£  +  2  8) 

30.  The  different  values  of  the  earth  pressure  determined 
by  graphical  computation  according  to  the  theory  of  Pro- 
fessor Rebhann  are  very  close  to  those  obtained  from  the 
formulas  of  Rankine  and  Weyrauch.  This  may  be  seen 
from  the  following  tables  prepared  from  the  class  work  by 
Mr.  W.  J.  Miller. 

In  the  following  tables  are  given  in  pounds  the  value  of  the 
earth  pressure  against  walls  of  different  heights.  The  walls 
were  considered  to  be  vertical,  leaning  5°  forward  and 
inclined  5°  toward  the  embankment.  In  the  first  table  the 
bank  of  earth  was  considered  with  no  surcharge,  or  when 
a  =  0  ;  in  the  second  table  when  the  surcharge  made  with 
the  horizontal  an  angle  of  10°,  or  a  =  10°.  Finally,  in  the 
last  table,  the  surcharge  was  considered  parallel  to  the  line 
of  natural  slope  of  the  material,  or  when  a  =  <f>.  In  every 
case  the  weight  of  the  earth  was  assumed  to  be  96  Ib.  per 
cubic  foot,  and  the  angle  of  natural  slope  (f>  =  30°. 


RETAINING  WALLS:    ANALYTICAL   METHODS         79 


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80       EARTH   SLOPES,   RETAINING  WALLS,   AND  DAMS 


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RETAINING  WALLS:   ANALYTICAL   METHODS 


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CHAPTER   IV 

THE   DESIGN   OF   RETAINING  WALLS 

31.  Types  of  Retaining  Walls.  —  Retaining  walls  are  built 
of  a  variety  of  sections.      For  convenience  these  may  be 
grouped  as  follows : 

I.    Plain  retaining  walls. 
II.    Retaining  walls  with  counterforts. 
III.    Retaining  walls  with  buttresses. 

Each  group  can  be  further  subdivided  into  many  varieties 
according  to  the  inclination  and  shapes  of  the  front  and  back 
of  the  wall. 

32.  Plain   Retaining   Walls.  —  The   plain   retaining   wall 
may  be  constructed  in  various  manners : 

(a)  With  vertical  front  arid  back,  as  indicated  in  Fig.  33. 
(5)  With  vertical  front  and  inclined  back,  as  indicated  in 
Fig.  34. 

(<?)  With  vertical  back  and  inclined  front,  Fig.  35. 

(d)  With  both  front  and  back  inclined,  but  in  opposite 
directions,  as  indicated  in  Fig.  36. 

(e)  With  front  and  back  inclined  in  the  same  direction, 
Fig.  37. 

(/)  With  stepped  back.  In  order  to  make  the  wall  thicker 
toward  the  bottom,  the  back  of  the  wall  may  be  stepped 
while  the  front  is  plain,  either  vertical  (Fig.  38)  or  inclined 
(Fig.  39).  The  back  of  the  wall  in  the  successive  steps  may 
be  either  vertical  or  inclined. 

82 


A  ^  YS* 

OF  THE     ' 

UNIVERSITY 


THE   DESIGN  OF  RETAINING  WALLS 


83 


With  curved  face.  English  engineers  often  make  the 
back  of  the  wall  vertical,  whether  stepped  or  not,  while  the 
front  of  the  wall  is  curved  to  a  circular  arc  of  radius  gener- 


ally equal  to  twice  the  height  of  the  wall,  the  center  of  the 
arc  being  located  on  the  line  of  the  top  of  the  wall  produced. 
Such  a  section  is  shown  in  Fig.  40. 

(h)  With  curvilinear  back  and  front.  German  engineers 
often  make  the  front  of  the  retaining  wall  circular,  follow- 
ing the  English  manner,  but  at  the  same  time  make  the  back 
of  the  wall,  which  is  usually  of  stepped  construction,  also 
circular  and  parallel  to  the  front,  as  indicated  in  Fig.  41. 

(i)  French  engineers  have  adopted  a  profile  composed  of 


rectilinear  segments  arranged  with  different  inclinations  so 
that  the  wall  approximates  a  curvilinear  outline.  Begin- 
ning at  the  top  of  the  wall,  for  the  first  3  yards  they  make 


84       EARTH   SLOPES,  RETAINING  WALLS,   AND  DAMS 


FIG.  42. 


the  inclination  of  ^,  for  the  second  3  yards,  ^,  and  for  the 
residual  height,  no  matter  how  high  the  wall,  the  constant 

inclination  of  T3^  of  the  height. 

The     back     of     the     wall     is 

formed  with  steps  about  1  ft. 

wide  for  every  5  ft.  in  height. 

The  French  profile  is  shown  in 

Fig.  42. 

33.  Retaining  Walls  with 
Counterforts.  —  Retaining  walls 
with  counterforts  are  similar 
to  plain  retaining  walls  as 
described  above,  but  along  the 

back  of   the  wall,  at   intervals,  are  constructed  blocks  of 

masonry  of  larger  dimensions  reenforcing  the  wall  proper. 

These  blocks  of  masonry  may  be 

constructed  on  either  rectangular 

or    trapezoidal     ground    plan,    as 

indicated  in  Figs.  42  and  43,  and 

they  may  be  vertical  or  inclined 

in  profile.     They  may  even    have 

a  stepped  outline.  Such  counter- 
forts are  located  usually  from  12 

to  20  ft.  apart,  center  to  center. 

34.  Retaining  Walls  with  But- 
tresses. —  Retaining  walls  with 
buttresses  (Figs.  45  and  46)  are 
very  similar  to  those  with  coun- 
terforts, the  only  difference  being  that  the  pillars  or  pro- 
jecting blocks  of  masonry  are  on  the  front  of  the  wall 
instead  of  on  the  back.  The  buttresses  may  be  rectangular 


\ 

r 

*— 

«w 

1 

FIG.  43. 


THE   DESIGN   OF  RETAINING   WALLS 


85 


w 

m 

I 

jr 
'.             ^m 

aw 

1 

or  trapezoidal  in  plan,  and 
either  vertical  or  inclined  in 
profile,  according  to  the  shape 
of  the  front  of  the  wall,  but 
they  are  never  recessed.  But- 
tresses are  spaced  about  the 
same  as  counterforts,  viz.,  from 
12  to  20  ft.,  center  to  center. 

In  very  heavy  soils  to  re- 
lieve the  pressure  of  the  earth 
against  the  retaining  wall,  hori- 
zontal arches  sprung  between 
the  counterforts  may  be  used. 

In  such  cases  the  counterforts   must    be    designed    strong 

enough    to    resist    the 

pressure  which  is  trans- 
mitted to  them  by  the 

arches.         The     arches 

may  .be    built    in    one 

row  or  in  several  rows, 

and     the    lowest    may 

sometimes  be  inverted, 

as  in  Fig.  47. 


FIG.  45. 


FIG.  46. 


FIG.  47. 

THE  EQUILIBRIUM  OF  RETAINING  WALLS 

35.  The  forces  acting  on  a  retaining  wall  are  the  pressure 
of  the  earth  P  and  the  weight  of  the  wall  W.  It  has  been 
demonstrated  that  the  point  of  application  of  the  earth  pres- 
sure is  at  a  distance  above  the  base  of  one  third  the  height 
of  the  wall.  The  weight  of  the  wall  is  assumed  to  be  con- 
centrated at  its  center  of  gravity. 


86        EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 


The  pressure  P  (Fig.  48)  being  inclined,  making  an 
angle  <£'  with  the  normal  to  the  back  of  the  wall,  may  be 
resolved  into  its  vertical  and  horizontal  components.  The 
horizontal  component  thrusts  against 
the  wall,  tending  to  produce  both 
sliding  and  rotation  ;  the  vertical 
component,  on  the  other  hand,  assists 
the  weight  W  in  pressing  down  the 
*  different  strata  of  the  wall  and 
thereby  counteracting  the  overturn- 
ing and  sliding  tendency  of  the 
horizontal  thrust.  Under  the  action 
of  these  forces  the  wall  tends  to 
overturn  in  whole  or  in  part.  The 
most  dangerous  section  is  usually  the  section  at  the  base 
of  the  walls,  since  the  earth  pressure  increases  as  the 
square  of  the  height  and  therefore  increases  faster  than 
the  resisting  forces,  the  weight  of  the  wall.  The  ten- 
dency of  the  resultant  R  of  the  forces  P  and  W 
to  approach  the  exterior  of  the  wall  is  greatest  at  the 
base.  For  this  reason  the  stability  of  the  wall  is  always 
determined  with  respect  to  the  section  at  the  base,  and 
the  thickness  of  any  wall  to  resist  a  given  pressure  is 
always  calculated  for  the  base  of  the  wall. 

As  a  rule,  a  retaining  wall  is  in  less  danger  of  failure  by 
sliding  than  failure  by  overturning  ;  consequently  when  a 
wall  is  safe  against  overturning  it  is  generally  safe  against 
sliding. 

In  determining  the  thickness  of  retaining  walls  it  is  gen- 
erally assumed  that  the  wall  is  laid  up  dry,  i.e.  without 
mortar,  resisting  only  by  virtue  of  its  own  weight,  and 
having  no  tensile  strength.  Tensile  forces  are  produced  in 


THE   DESIGN   OF   RETAINING  WALLS  87 

a  wall  when  the  pressure  is  so  near  one  edge  of  the  wall 
and  is  so  great  that  it  tends  to  force  down  the  material  at 
this  edge  and  raise  up  or  tear  apart  the  material  at  the  other 
edge. 

In  order  that  a  wall  may  be  strong  enough  to  resist  the 
action  of  the  forces  P  and  Win  all  the  probable  sections  of 
rupture,  the  following  conditions  must  be  satisfied : 

1.  The   resultant   R  of   the  forces  P  and    W  must  fall 
within  the  middle  third  of  the  width  of  the  base.     This  con- 
dition, while  it  makes  the  wall  safe  against  overturning,  also 
makes  tensile  stresses  impossible. 

2.  The  direction  of  the  resultant  R  must  make  a  smaller 
angle  with  the  normal  to  the  section  than  the  angle  of  fric- 
tion of  the  masonry  of  the  wall.     This   condition  insures 
stability  against  failure  by  outward  sliding. 

3.  The  greatest  compressive  stress  must  be  smaller  than 
the  crushing  strength  of  the  material  composing  the  wall. 
This  condition  insures  stability  against  crushing. 

The  stability  of  a  retaining  wall  may  be  calculated  in  two 
different  ways :  either  fixing  the  condition  of  equilibrium  of 
the  wall  and  adding  a  factor  of  safety,  or  requiring  as  a 
condition  of  safety  that  the  resultant  R  shall  pass  through 
the  middle  third  of  the  base. 

For  calculations  according  to  the  first  method,  let  m  =  the 
lever  arm  of  P,  then  Pm  will  be  the  product  of  P  by  its 
distance  from  the  foot  of  the  wall  A,  w  =  the  lever  arm 
of  TF,  and  Ww  the  product  of  W  by  its  distance  from  A. 

The  wall  will  be  in  equilibrium  when 

Pm  =  Ww. 

The  wall  is  safe  if  it  is  able  to  resist  greater  pressure  than 
P.  It  is  usual  for  this  purpose  to  multiply  the  moment  of 


88        EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 

P  by  a  factor  of  safety,  which  is  generally  fixed  somewhere 
between  1.50  and  2. 

Such  a  method,  although  giving  good  results,  especially 
when  a  factor  of  safety  of  2  is  used,  does  not  convey  any 
idea  of  the  distribution  of  stresses  in  the  masonry.  Thus, 
for  example,  a  wall  may  be  safe  against  overturning,  being 
made  to  fulfill  the  condition  Pm  =  Ww  with  a  factor  of  safety 
of  2  and  yet  it  may  crack  and  be  ruined  because  the  compres- 
sive  stresses  in  the  wall  are  concentrated  on  a  small  portion 
instead  of  being  evenly  distributed,  and  because  tensile 
stresses  are  thus  set  up. 

The  second  method  is  the  more  rational  one ;  and  when 
the  engineer  has  given  such  proportions  to  the  wall  that  the 
compressive  stress  remains  below  the  crushing  strength  of 
the  material,  he  may  depend  upon  the  stability  of  his  wall. 

DETERMINATION  OF  WIDTH  OF  BASE  BY  GRAPHICAL 
METHODS 

36.  In  the  previous  chapters,  in  calculating  the  pressure 
of  the  earth  against  a  retaining  wall,  it  has  uniformly  been 
assumed  that  the  length  of  wall  considered  was  one  unit, 
consequently  that  the  wedge  which  tends  to  slide  down 
and  thus  causes  the  thrust  against  the  back  of  the  wall  had 
a  length  of  one  unit.  For  this  reason  we  have  always  used 
the  area  of  the  triangle  of  the  base  as  equivalent  to  the 
volume  of  the  wedge  itself.  Continuing  this  practice,  we 
will  deal  with  a  retaining  wall  having  a  length  of  one  unit, 
perpendicular  to  the  plane  of  the  cross-section  (the  plane 
of  the  paper).  Then  all  calculations  can  be  referred  to  the 
profile  of  the  wall,  since  the  area  of  the  profile  is  numerically 
equal  to  the  volume  of  a  wall  one  unit  long. 


THE  DESIGN  OF  RETAINING   WALLS 


89 


From  what   has  gone  before,  the  pressure  of  the  earth 
against  the  back  of  the  wall  is  fully  known  as  to  direction, 
magnitude,  and  position.     When  the  weights  of   both  the 
earth  and  the  masonry  per  unit  of 
volume    are    known,   it    becomes    a  &' 

Very  simple  matter  to  oppose  to 
the  prism  of  earth  causing  the  pres- 
sure a  suitable  resisting  section  of 
masonry. 

Let    ABC    (Fig.     49)    represent 
the  cross-section  of  the  active  prism   . 
of  earth,  and  denote  by  <ym  and  <yt  the 
weight  of  a  unit  of  volume  of  ma- 
sonry and  earth  respectively.     Then   for  equal  weight  we 
will  have, 


A 

FIG.  49. 


from  which 


AC' 


ACy* 


then  AB  C'  will  represent  a  mass  of  masonry  of  weight  equal 
to  the  pressure  of  the  earth. 

A  retaining  wall  will  be  in  equilibrium  when  the  moment 
of  the  weight  is  equal  to  the  moment  of  the  pressure.  To 
design  a  safe  wall  it  is  necessary  to  multiply  the  moment  of 
pressure  by  a  certain  coefficient  m,  called  factor  of  safety; 
then  we  will  have 

Ww  =  Ppm, 

the  coefficient  m  is  usually  made  equal  to  1.5,  2,  or  even  3. 

37.  EXAMPLE  1.  Retaining  Wall  with  Vertical  Front  and 
Back.  —  Let  AB  (Fig.  50)  represent  the  back  of  the  wall. 
From  A  and  B  draw  the  lines  AX,  BX'  horizontally  ;  the 


90       EARTH  SLOPES,   HETAINING  WALLS,  AND   DAMS 


thickness  of  the  wall  is  supposed  to  have  been  fixed  by 
successive  trials.  Along  the  line  AX  lay  off  a  segment 
A\  equal  to  the  unit  of  length,  though  it  can  be  taken  of 
any  length,  and  erect  a  perpendicular,  giving  the  first  part  of 
cross-section  of  the  wall.  Draw  the  force  polygon,  mak- 
ing the  ray  o  0  equal  twice  the  earth  pressure,  or  oO  =  2  P 
and  parallel  to  the  direction  of  the  pressure  itself,  i.e.  making 
an  angle  $'  with  the  normal  to  the  back  of  the  wall.  In  the 
present  case,  we  will  assume,  it  is  proposed  to  build  a  wall 

capable  of  resisting  twice  the 
pressure  of  the  earth,  that  is, 
a  wall  having  a  factor  of 
safety  of  2.  For  this  reason 
we  made  oO—^P.  From  o 
draw  a  vertical  line  and  lay 
off  a  segment  01  equal  to  the 
weight  of  the  first  part  of 
the  wall.  Join  1  with  0\ 
this  will  be  the  resultant  of 
the  total  pressure  combined 
with  the  weight  of  this  part 
of  the  wall.  Then  at  one  third  of  AB  draw  the  line  of 
the  earth  pressure,  produce  it  until  it  strikes  at  I  the 
center  line  of  this  section  of  the  wall,  and  draw  from  /a  line 
parallel  to  the  resultant  10. 

Again,  along  AX  take  a  second  segment,  and  erect  a 
perpendicular.  Find  the  center  of  this  second  section  of 
wall  and  produce  the  first  resultant  until  it  meets  this 
•center  line  at  II.  In  the  force  polygon  lay  off  a  segment 
1-2  equal  to  the  weight  of  this  second  prism  of  wall. 
Draw  the  resultant  02,  and  from  II  draw  a  parallel  to  this 
resultant.  Proceed  in  the  same  manner  by  adding  continu- 


FIG.  50. 


THE  DESIGN  OF   RETAINING  WALLS 


91 


ously  small  sections  of  wall,  until  a  point  is  reached  where 
the  last  resultant  strikes  the  base  line  AX  at  O.  Erect  the 
perpendicular  CD,  the  wall  ABOD  will  be  the  required  wall, 
being  in  equilibrium  under 
twice  the  actual  earth 
pressure. 

38.  EXAMPLE  2.  Retain- 
ing Wall  with  Vertical  Front 
and  Inclined  Back.  —  Let  AB 

be  the  back  of  the  wall  (Fig. 
51).  Draw  the  horizontals 
AX  and  BX'.  From  the 
point  B  drop  the  perpendicu- 


lar BB'  and  divide  the  unde- 


jr 


termined  rectangular  portion 
to  the  left  of  BB'  into  a  number  of  small  vertical  stripes 
as  above.  Draw  the  force  polygon  to  such  a  scale  that  the 
segment  o\  will  represent  the  weight  of  the  triangular  prism 
BBJ A  and  the  segments  1.2,  2.3,  3.4,  etc.,  the  corresponding 
weights  of  the  successive  rectangular  prisms  of  masonry. 
The  construction  will  be  perfectly  similar  to  the  one  given 
in  the  case  above  except  that  the  earth-pressure  line  is  con- 
tinued to  the  vertical  through  the  center  of  gravity  of  tri- 
angle ABB'.  Where  the  polygon  of  the  resultant 
pressure  meets  the  line  AX  draw  the  perpendicular  (71), 
the  figure  CDAB  will  be  the  required  section  of  the 
wall. 

39.  EXAMPLE  3.  Retaining  Wall  with  Inclined  Front  and 
Vertical  Back.  —  In  Fig.  52,  AB  represents  the  back  of  the 
wall  and  CD  the  given  inclination  of  the  front.  Draw 
the  usual  horizontal  lines,  and  from  B  draw  B\  parallel 


92        EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 


to  CD.  Along  the  upper  line  mark  off  the  points  .F,,2,  etc.,  and 
from  these  draw  lines  parallel  to  CD,  the  intervals  will  represent 
so  many  prismoids  of  masonry.  Then  draw  the  force  poly- 
gon, and  along  the  line  of  weights  lay  off  the  first  segment, 
01,  to  represent  the  weight  of  the  triangular  prism  AB1, 

and  then  in  succession  the 
weights  of  the  prismoidal  strips 
lying  to  the  left  of  Bl.  Find 
the  center  of  gravity  of  the 
various  prisms  of  masonry  and 
draw  the  resultants  in  the  usual 
manner,  continuing  them  in 
each  case  to  intersect  the  verti- 
cals through  the  centers  of 
gravity.  At  the  point  E,  where 
FIG.  52.  the  polygon  of  resultants  strikes 

the  base  line,  draw  the  line  EF  parallel  to  the  given  in- 
clination CD.  The  figure  ABEF  will  be  the  cross-section 


PB 


of  the  required  wall  with  the, 
face  parallel  to  the  given  in- 
clination. 

40.  EXAMPLE  4.  Retaining 
Wall  with  Faces  inclined  in 
Opposite  Directions.  —  The  line 
AB  (Fig.  53)  represents  the 
back  of  the  wall.  From  B 
draw  a  perpendicular  and 
treat  the  first  triangle  as  in 
Example  2.  Then  deal  with 
the  rest  of  the  wall,  up  to  the  front  face  CD,  according  to 
the  method  followed  in  Example  3.  The  resulting  profile 
ABCD  is  the  cross-section  of  the  wall. 


FIG.  53. 


THE   DESIGN  OF  RETAINING  WALLS 


93 


FIG.  54. 


41.  EXAMPLE  5.     Retaining  Wall  with  Parallel  Inclined 
Faces.  —  The  construction  (Fig.  54)  is  perfectly  similar  to 
that  of  Example  1,  being  care- 
ful   to    use    as  vertical    center 

lines  the  verticals  through  the 
centers  of  gravity  of  the  several 
sections  of  the  wall.  When 
both  faces  are  inclined  in  the 
same  direction  but  are  not 
parallel,  the  general  method  of 
Example  4  is  to  be  used. 

42.  Interpolation  Method.  — 
All    the    above    problems    can 

be  solved  in  another  way,  employing  the  principle  that  the 
required  section  of  the  wall,  which  is  in  equilibrium  with 
the  pressure  (1.5,  2,  or  3  times  the  actual  pressure,  according 
to  the  factor  of  safety),  is  included  between  two  cross- 
sections  of  which  one  has  excessive  stability  and  the  other 
deficient  stability.  In  other  words,  we  first  determine  a 
section  of  wall  in  which  the  resultant  between  the  pressure 
and  weight  falls  outside  the  base  of  the  section;  it  is  evi- 
dent that  such  a  wall  will  not  be  strong  enough  to  resist 
the  assumed  pressure.  Then  we  determine  a  section  of  wall 
in  which  the  resultant  falls  within  the  base ;  such  a  wall  will 
be  too  strong.  The  required  cross-section  will  be  between 
these  two. 

In  Fig.  55,  AECD  and  ABEF  are  the  two  cross-sections 
thus  determined,  and  R  and  Rl  their  respective  resultants. 
From  C  drop  a  perpendicular  and  lay  off  a  downward  seg- 
ment Cm  =  HO.  At  E  erect  a  perpendicular  and  lay  off 
upward  a  segment  En  =  ERf.  Connect  n  with  m  ;  at  the 
point  of  intersection  with  the  ground  line  draw  a  line  paral- 


94       EARTH   SLOPES,   RETAINING   WALLS,   AND    DAMS 


FIG.  55. 


lei  to  CD.  The  figure  ABCrH  will  be  the  required  section 
of  the  wall.  In  this  construction  it  is  assumed  that  as  the 
line  representing  the  front  of  the  wall  moves  from  CD  to 

EF,  the  curves  traced  by  points 
located  in  the  manner  m  and  n 
were  located  in  a  straight  line. 
Really  it  is  a  curve  of  small 
curvature,  but  the  error  in  as- 
suming it  to  be  straight  is  so 
small  as  to  be  negligible  in 
practical  work. 

It  may  be  required  that  the 
resultant  of  the  actual  earth 
pressure  (not  multiplied  by  a 
factor  of  safety)  and  the  height 
of  the  wall  should  fall  within  the  middle  third  of  the 
base.  In  this  case  proceed  in  the  following  manner  :  after 
the  thickness  of  the  wall  has 
been  determined  by  any  of  the 
methods  already  explained,  in 
the  force  polygon  (Fig.  56), 
draw  OS  equal  to  the  actual 
earth  pressure  and  weight  of 
the  wall.  On  the  back  of  the 
wall  AB  produce  the  line  of 
pressure  to  intersect  at  Q  the 
vertical  dropped  from  the  cen- 
ter of  gravity  of  the  figure 
AB&H.  From  Q  draw  a  line 
parallel  to  the  resultant  SM,  intersecting  the  base  AG- 
at  N.  Now  if  this  resultant  falls  either  within  or  just  at 
the  extreme  end  of  the  middle  third  of  the  base,  the  wall 


H 


6/V 


FIG.  56. 


THE   DESIGN   OF   RETAINING  WALLS 


95 


satisfies  the  given  condition.  But  if  it  falls  outside  the 
middle  third,  the  thickness  of  the  wall  must  be  increased 
until  the  new  resultant  does  fall  inside  the  middle 
third. 

43.  EXAMPLE  6.  Retaining  Walls  with  Counterforts.  - 
The  graphical  determination  of  retaining  walls  with  coun- 
terforts and  buttresses  becomes  very  simple  when  such  walls 
are  considered  as  composed  of  two  parts  having  different 
specific  weights.  The  stability  of  the  wall  will  then  be 
determined  per  unit  of  length  just  as  in  the  cases  already 
reviewed.  For  the  sake  of  simplicity  in  the  graphical 
construction  it  is  assumed  that  both  counterforts  and  but- 
tresses are  located  12  ft.  apart,  center  to  center,  and  that 
they  are  3  ft.  wide. 

Under  retaining  walls  with  counterforts,  two  cases  may 
be  considered  :  (1)  the  thickness  of  the  wall  being  given, 
it  is  required  to  determine  the  thickness  of  the  counter- 
fort ;  or  (2),  the  thickness  of  the  counterforts  being  given, 
the  thickness  of  the  wall  is  to  be  determined.  We  will 
begin  with  the  first  case,  and  we  will  consider  the  condi- 
tion when  both  wall  and  counterforts  are  vertical. 

Referring  to  Fig.  57,  AB 
is  the  given  thickness  of  the 
wall.  It  is  required  to  de- 
termine the  thickness  AE  of 
the  counterfort.  This  is 
done  by  successive  trials,  add- 
ing small  slices  of  wall,  just 
as  in  the  examples  already 
discussed.  The  pressure  P 
divides  in  the  proportion  of  f 
on  the  wall  and  on  the  FIG.  57. 


96        EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 

counterfort.  Draw  the  force  polygon  in  the  usual  manner. 
The  pressure  on  the  counterfort  will  be  represented  by 
00;  the  weights  of  the  various  slices  of  the  counterfort 
should  be  taken  at  J  the  weight  of  equal  sections  of 
wall.  Suppose  AE  is  the  first  size  of  counterfort  we 
try.  Its  weight  will  be  represented  by  01  in  the  force 
polygon;  draw  the  line  1.2  parallel  to  the  pressure  and 
equal  to  |  P  (or  1 2  P  in  case  2  is  used  as  factor  of 
safety).  Draw  a  perpendicular  and  lay  off  a  segment 
2.3  equal  to  the  weight  of  the  given  wall ;  the  line  3.0 
will  be  the  resultant.  In  the  figure  draw  the  line  of  pres- 
sure at  ^E1FV  produce  it  to  meet  the  vertical  from  the 
center  of  gravity  of  the  prism  of  masonry  ABE^F^  and 
from  this  point  draw  a  line  parallel  to  the  resultant  03, 
until  meeting  at  M  the  line  of  pressure  which  strikes  the 
back  of  the  wall  AB  at  ^  the  height  from  A.  Draw  from 
M  a  parallel  to  the  direction  of  pressure  02  of  the  force 
polygon,  meeting  at  S  the  vertical  dropped  from  the  center 
of  gravity  of  the  wall  ABCD,  and  from  8  draw  a  parallel 
to  the  resultant  03,  meeting  the  ground  line  at  the  point  av 
Take  a  second  segment  E^F-^E^F^  add  its  weight  l.lj  in 
the  force  polygon  arid  proceed  as  above ;  we  then  find 
that  the  last  resultant  drawn  from  S1  meets  the  ground  line 
at  the  point  #2.  At  aA  and  #2  erect  vertical  lines  and  lay 
off  segments  «151  and  a2b2  equal  to  AEl  and  AE^  respec- 
tively ;  connect  b1  and  b2  and  produce  this  line  until  it 
intersects  the  line  of  the  front  of  the  wall  at  d.  The 
length  of  the  line  Cd  represents  the  required  thickness 
of  the  counterfort. 

The  prism  of  masonry  of  the  counterfort  ABEF  dis- 
places a  similar  prism  of  earth  which  was  previously  cal- 
culated as  exerting  thrust  against  the  wall.  The  actual 


THE  DESIGN  OF   RETAINING  WALLS 


97 


HFB 


FIG.  58. 


pressure   against   the  counterforted  wall  will   therefore  be 
a  little  smaller  than  the  value  used. 

(5)  When  the  thickness  of  the  counterfort  ABCD  is 
given  and  it  is  required  to  determine  the  thickness  of 
the  wall,  the  problem  is  easily 
solved.  Draw  the  force  poly- 
gon in  Fig.  58  by  laying  off 
in  the  direction  of  the  pres- 
sure a  segment  mo  equal  to 
\  P  (or  \  2  P  in  case  a  factor 
of  safety  2  is  used) ;  then 
draw  a  vertical  line  and  lay 
off  a  segment  mn  equal  to 
the  weight  of  the  counter- 
fort, and  from  n  in  the  direc- 
tion of  the  pressure  draw  nr 
equal  f  P  (or  |2  P).  From  the  point  r  draw  a  vertical 
line  and  lay  off  segments  equal  to  the  weights  of  the 
various  slices  of  masonry  ABEF,  EFG-H,  etc.,  and  pro- 
ceed as  in  Example  1. 

44.  EXAMPLE  7.  Retain- 
ing Wall  with  Buttresses. — 
The  thickness  of  the  wall 
ABCD  (Fig.  59)  being  given, 
that  of  the  buttress  ABEF 
is  easily  determined  as  fol- 
lows: draw  the  force  poly- 
gon in  the  usual  way,  making 
oO  equal  to  twice  the  actual 
earth  pressure ;  then  lay  off 
vertically  a  segment  0!  equal  to  the  weight  of  the  wall 
ABCD.  Try  a  buttress  of  thickness  AM  and  lay  off  in  the 


FN  B 


FIG.  59. 


98     EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 


K 


force  polygon  a  segment  1-2  equal  to  one  fourth  the  weight 
of  the  prism  AMNB;  draw  the  resultant  2.0.  In  the  figure 
draw  the  pressure  polygon  and  if  the  resultant  strikes  outside 
the  base,  try  a  new  thickness  of  buttress.  When  the  final  re- 
sultant meets  the  ground  line  at  the  edge  of  or  just  within 
the  base  of  the  buttress  the  required  thickness  AE  of  the 
buttress  ABEF  will  have  been  found. 

45.  EXAMPLE  8.  Retaining  Wall  with  Inclined  But- 
tresses. —  In  case  of  inclined  buttresses,  the  necessary  thick- 
ness at  the  base  is  determined  by  trial.  When  the  wall 

ABCD  (Fig.  60)  is  given, 
the  force  polygon  001  can 
be  easily  drawn ;  the  center 
of  pressure  or  resultant  will 
certainly  fall  outside  the  base 
of  the  wall,  otherwise  there  is 
no  need  of  buttresses.  As- 
sume, then,  any  thickness  EC 
for  the  buttresses  and  draw 
from  E  a  line  parallel  to  the 
given  inclination  IK\  lay  off 
in  the  force  polygon  a  seg- 
ment 1,  2  equal  to  one  fourth  the  weight  of  the  prism  CDEF. 
From  the  point  where  the  resultant  emerging  from  the  wall 
meets  a  vertical  dropped  from  the  center  of  gravity  of  CDEF, 
draw  a  line  parallel  to  02  of  the  force  polygon.  This  will 
fall  either  within  or  without  the  wall ;  in  the  former  case 
there  is  an  excess  of  stability,  and  the  thickness  of  the 
buttresses  can  be  taken  as  EC  or  reduced  by  a  new  trial, 
while  in  the  latter  case  it  will  be  necessary  to  increase  the 
thickness  of  the  base,  using  the  methods  already  indicated 
for  walls. 


FIG.  60. 


THE   DESIGN  OF   RETAINING  WALLS  99 

46.  EXAMPLE  9.  Retaining  Walls  with  Relieving  Arches 
between  Counterforts.  —  In  this  case  there  will  be  back  of 
the  wall,  besides  the  counterforts,  the  masonry  arches  and 
the  earth  packed  between  the  arches.  In  order  to  deter- 
mine the  thickness  of  the  counterforts  and  arches,  it  is 
necessary  to  determine  the  average  specific  weight  of  the 
structures  and  materials  back  of  the  wall. 

Let  A  =  the  area   included   between   the   center  lines  of 

two  consecutive  counterforts, 
Al  =  the    area   occupied   by   the   earth    between    the 

arches, 

Al  —  the  weight  of  the  earth  per  cubic  foot, 
A%  =  the  area  occupied  by  masonry  arches  and  counter- 

forts, and 

A2  =  the  weight  of  the  masonry  per  cubic  foot. 
The  average  specific  weight  back  of  the  wall  will  be 


A 

After  the  unit  of  weight  has  been  determined,  the  thickness 
either  of  the  counterforts  when  the  wall  is  given,  or  of  the 
wall  when  the  counterforts  are  given,  is  obtained  as  in  the 
case  of  simple  counterforts  indicated  above. 

In  the  case  of  a  retaining  wall  with  relieving  arches 
between  the  buttresses,  the  solution  is  similar  to  that 
indicated  in  Example  7  for  determining  the  thickness  of  a 
buttress.  In  that  example  the  buttresses  taken  as  weighing 
only  one  fourth  of  the  weight  of  the  masonry  of  the  wall, 
because  they  were  only  1  yard  wide  and  4  yards  apart. 
In  the  present  case,  however,  the  average  specific  weight  of 
the  buttresses  and  arches  is  to  be  calculated  as 


100      EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 


When  this  value  has  been  determined  the  process  is  quite 
the  same  as  already  explained,  and  the  required  thickness  of 
buttress  is  found  directly. 

DETERMINATION  OF  WIDTH  OF   BASE  BY  ANALYTICAL 

METHOD 

47.  Suppose  ABCD  (Fig.  61)  to  represent  the  cross-sec- 
tion of  a  retaining  wall  having  inclined  front  and  vertical 
back.  The  forces  acting  on  the  wall  are  : 

(a)  The  outward  pressure  of  the  earth  backing,  P. 
(6)  The  weight  of  the  wall,  W. 

(<?)  The  reaction  of  the  foundation  upward  against  the 
base  AQ  of  the  wall,  which  is  equal  and  opposite  to  the 
resultant  R  of  the  two  forces,  P  and  W.  This  reaction  will 

be  denoted  by  R9. 

Resolve  the  forces  P  and  R  into 
their  horizontal  and  vertical  com- 
ponents;  call  T  and  V  the  hori- 
zontal and  vertical  components  of  the 
pressure  P,  and  T  and  V  the  hori- 
zontal and  vertical  components  of 
the  reaction  Rf . 

When  the  wall  is  in  equilibrium, 
the  algebraic  sum  of  the  forces  as 
FIG.  61.  well  as  the  algebraic  sum  of  their 

moments  must  be  zero.     Therefore 


W+  V-  V  =  0, 

and  Tt+Ww-  V'v  =  0, 

t,  w,  and  v  being  the  lever  arms  of  the  forces  T,  W,  and  F7, 


THE  DESIGN  OF   RETAINING  WALLS  101 

or  the  distances  of  their  points  of  application  from  A,  the 
interior  edge  of  the  wall. 

From  the  above  equations  we  have  : 

T=T', 

w+  v=  v1, 

Tt+Ww=Vfv. 

The  last  equation  gives  directly  the  value  of  v, 

Tt+  Ww 

~vr~ 

Substituting  to  V  its  value  W+  V,  we  obtain 


TF  +  V 

The  pressure  P  makes  an  angle  <j>f  with  the  normal  to  the 
back  of  the  wall,  and  in  the  present  case  with  the  horizontal. 
Since  jT,  the  projection  of  P  on  the  horizontal,  we  have 


Substituting  these  values  in  equation  (1),  there  results 
=  Pt  cos  <'  +  Ww 


If  from  A  we  draw  a  line  perpendicular  to  the  direction 
of  P,  its  angle  with  AB  will  be  equal  to  <ft  ;  the  length  of 
this  perpendicular  will  be  t  cos  <£/,  so  that  the  latter  (erm  in 
equation  (2)  may  be  replaced  by  m.  Then 

Pm  -h  Ww  /Q, 

v  =  -  •  (o) 

TP+Psin^'  '> 

In  order  that  the  resultant  R  fall  within  the  middle  third 
of  the  base,  v  must  lie  between  and  the  width  of  the 


102      EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 

base.     Calling  the  base  f,  it  will  have  the  smallest  allowable 
value  when 


or 


Pm+  Ww 


(4) 


which  is  the  general  formula. 

Assume  the  thickness  of  the  wall  at  the  top  to  be  0.6  of 
the  width  of  the  base  (in  practice  it  would  rarely  be  made 
smaller).  Then  the  weight  IF  of  the  trapezoidal  section  of. 
the  wall  is  expressed  by  the  formula 


Substituting  this  value  in  equation  (4)  and  solving  for  £, 
the  thickness  of  the  wall  at  the  base  is  determined  directly, 
without  resort  to  trial  calculation. 

The  case  considered  is  only  a  special  case,  however,  since 
we  assumed  that  the  back  of  the  wall  is  vertical.  The 
general  case  is  that  in  which  both  back  and  front  of  the 

wall  are  inclined.  Let  us  analyze 
this  case,  assuming  further  that  to 
increase  the  safety  against  sliding, 
the  base  of  the  wall  is  inclined  at 
an  angle  a  to  the  horizontal.  The 
angle  between  the  base  and  the 
back  of  the  wall  will  be  called  ft 
(Fig.  62). 

The  external  forces  are  the  same 
as  before  ;  namely,  P,  TFJ  and  R1  . 
Resolve  each  one  into  its  components, 
normal  and  parallel  to  the  base  A  O. 


THE   DESIGN  OF  RETAINING  WALLS 
The  forces  being  in  equilibrium,  we  have 


103 


V+  ITa-Fi  =  0, 
Pm  +  Ww  -  Vjv  =  0, 
from  which  we  deduce 

T=  T 


The  last  of  these  gives 


<p  =  Pm  -f  Ww. 


Pm  +  Ww 


Substitute  for  Vl  its  value  F-f  Wv  we  get 

Pm  + 


v  = 


F  + 


Now  the  pressure  P  (Fig.  63)  makes  an  angle  <£'  with  the 
normal  to  the  back  of  the  wall,  and  consequently  it  makes 
an  angle  (90  -  <£')  with  the  surface  AB.     Then  T  makes 
with  P  an  angle    180  —  /3— (90° 
-</>')  =  90°  -(£  -0'),    so     that 
the  angle  between  P  and  its  ver- 
tical   component    V  is    equal    to 

therefore 


The  vertical  component  F  of  the 
pressure  P  is  therefore  expressed 

by 

F=  P  cos  C/3  -  </>) 

=  P  cos  (90  -(«  -f-  3  +  <£')), 
or 

T^=P  sin  (a  +5  +  <£')• 


FIG.  63. 


104     EARTH   SLOPES,  RETAINING   WALLS,   AND   DAMS 

The  angle  between  W2  and  W  is  equal  to  angle  a,  since  the 
iwo  angles  have  sides  respectively  perpendiculars  ;  hence 

TF2  =  TFcos  a. 
On  substituting  these  values  of    V  and   W2  in  equation  5, 

•we  find 

Ww 


_ 
"  JFcos  «  +  P  sin  («  +  8  +  <£') 

In  order  that  the  resultant  shall  pass  through  the  outer 
•edge  of  the  middle  third  of  the  base,  the  base  of  the  wall 
must  be  given  a  width  of  t  =  f  v,  or 

_  3  Pm  +  Ww 

v  —   ~^ 


2  IF  cos  a +  P  sin  («  + 

On  the  other  hand,  if  it  is  required  that  the  wall  shall  be 
able  to  resist  the  pressure  P  with  a  factor  of  safety  of  n 
against  overturning,  we  substitute  Pn  for  P  and  make  the 
base  of  the  wall  just  wide  enough  so  that  the  resultant 
strikes  the  outer  edge  of  the  wall ;  that  is,  we  make  t  =  v. 
In  this  case  the  formula  becomes, 

nPm+  Ww 

TFcosa  +  Psin  (a  +  S  +  6') 


CHAPTER  V 

DAMS 

48.  Kinds  of  Dams.  —  Walls  intended  to  retain  water 
instead  of  earth  are  called  udams."  They  can  be  consid- 
ered as  masonry  structures  erected  for  the  purpose  of  rais- 
ing the  level  of  water. 

Dams  may  be  grouped  in  two  classes:  ordinary  dams 
and  submerged  dams.  Ordinary  dams  are  those  to  retain 
or  store  water,  forming  a  reservoir  or  lake.  They  are 
higher  than  the  level  of  the  water  in  the  reservoir  or  lake. 
Submerged  dams,  on  the  other  hand,  have  the  object  of 
raising  the  level  of  water  in  a  river ;  since  the  water  will 
continue  to  flow  down  the  channel  it  rises  sufficiently  to 
overflow  the  structure,  so  that  the  dam  will  be  entirely 
under  water  or  submerged. 

In  regard  to  their  manner  of  resisting  the  pressure  of 
the  water,  dams  may  be  classed  as  gravity  dams  and 
arched  dams.  Gravity  dams  are  those  which  resist  the 
pressure  exclusively  by  their  own  weight.  Arched  dams, 
however,  are  those  in  which  the  pressure  is  transferred 
through  the  body  of  the  masonry  to  the  sides  of  the  valley, 
against  which  the  structure  abuts.  Arched  dams  are  con- 
structed in  the  shape  of  an  arch  having  the  two  side  hills 
for  abutments,  or  skewback. 

Another  type  of  dam  used  in  many  instances  is  the  res- 
ervoir embankment  or  earth  embankment.  These  usually 
contain  a  masonry  core  wall,  but  as  regards  their  static 
resistance  against  water  pressure  they  are  to  be  consid- 

105 


106    EARTH   SLOPES,  RETAINING  WALLS,   AND   DAMS 

ered  simple  earth  embankments,  since  the  core  wall  is  used 
only  to  prevent  seepage  of  water  through  the  embankment 
and  to  tie  the  mass  of  earth  together,  and  not  for  the  pur- 
pose of  resisting  the  pressure  of  water.  Earth  embank- 
ments are  not  considered  in  this  work. 

Ordinary  dams  act  exactly  like  retaining  walls  and  by 
the  use  of  the  principles  and  methods  already  given  it  will 
be  an  easy  matter  to  determine  their  proper  proportions  when 
we  know  the  direction,  magnitude,  and  point  of  application 
of  the  pressure  which  the  structure  is  required  to  resist. 

49.  Direction  of  the  Water  Pressure.  —  The   pressure  of 
water   is   always   perpendicular   to    the  back  of   the    dam. 
This  fact  was  demonstrated  by  Pascal  long  ago.     He  per- 
formed  an  experiment   in   which  he   forced   water   into   a 
hollow  metal  ball  pierced  with  very  small  'holes  at  various 
points.     The  water  squirted  out  of  all  the  holes  with  the 
same  velocity,  hence  the  same  pressure.     It  was  concluded 
that  since  the  pressure  in  the  water  was  equal  in  all  direc- 
tions, it  must  act  in  directions  perpendicular  to  the  surface 
inclosing  it. 

Referring  to  our  discussion  of  earth  pressure  against 
retaining  walls,  it  was  shown  that  the  direction  of  the 
pressure  makes  an  angle  <f>f  with  the  normal  to  the  back 
of  the  wall,  and  we  have  always  assumed  that  fi  =  <£,  the 
angle  of  natural  slope  of  the  material.  When  the  material 
is  water  instead  of  earth,  the  angle  of  natural  repose  is 
equal  to  zero,  so  that  fi  also  will  be  equal  to  zero,  and  con- 
sequently the  direction  of  the  pressure  will  coincide  with 
the  normal  to  the  back  of  the  wall. 

50.  Amount  of  the  Pressure.  —  In  determining  the  pres- 
sure of   earth  against  a  retaining  wall   graphically  it  was 
found  that  the  value  of  the  pressure  was  given  by  the  area 


DAMS 


107 


of  a  certain  triangle  multiplied  by  the  unit  of  weight  of 
the  material.  The  triangle  of  pressure  was  shown  to 
have  one  of  its  sides  parallel  to  the  directrix,  a  line 
drawn  so  as  to  make  the  angle  </>  4-  <£'  with  the  back  of 
the  wall,  and  another  side  lying  in  the  surface  of  repose, 
these  two  sides  being  equal  to  each  other.  Since  the  con- 
struction used  for  this  triangle  of  pressure  is  valid  for 
all  values  of  c/>  and  <£>',  independently  of  the  nature  of  the 
material,  it  will  apply  also  to  water,  a  material  in  which 
the  angle  of  natural  repose  is  equal  to  zero. 

Suppose  the  line  AB  (Fig.  64)  represents  the  back  of  a 
dam  and  the  line  DO  the  surface  of  water  in  the  reservoir 
back  of  the  dam.     Since 
both  <£  and  <£/  are  equal  ^ 

to  zero,   the    surface    of  , 

repose  will  coincide  with 
the  horizontal  line  AE, 
and  the  directrix  coin- 
cides with  the  back  of 
the  dam  AB.  The  tri- 
angle of  pressure  is  then 
easily  determined.  The  — — 
line  AD  will  be  the  side  > 

of  the  isosceles  triangle 
parallel  to  the  directrix,  and  by  measuring  a  length 
equal  to  AD  on  the  horizontal  line  AE,  the  surface  of 
repose,  we  determine  the  point  F.  Joining  D  with  F, 
there  results  the  triangle  of  pressure  ADF,  whose  area 
multiplied  by  the  specific  weight  of  water  will  give  the 
total  pressure  against  the  back  of  the  dam.  Thus, 


F 


FIG.  64. 


108     EARTH   SLOPES,   RETAINING  WALLS,   AND   DAMS 

which  can  be  written  also 

P  =  i  AD  x  AF  x  7. 

But  AJ)  —  AF=  h,  the  depth  of  water  in  the  reservoir;   sub- 
stituting this  value  gives 

P  =  ^h  x  h  x  y, 

or  P  =  I  h*y. 

which  is  the  well-known  formula  for  determining  the  hori- 
zontal pressure  of  water  against  a  dam. 

51.  Point  of  Application  of  the  Pressure.  —  The  point  of 
application  of  the  resultant  pressure  on  the  back  of  the  dam 
is  a  distance  above  the  base  equal  to  one  third  the  depth  of 
water. 

In  Fig.  65,  the  triangle  AB  C  represents  the  total  pressure 
of  the  water  when  the  depth  of  water  is  equal  to  AB.  For 

any  other  point,  D,  the  total 
pressure  on  that  part  of  the 
dam  above  the  point  will  be 
given  by  the  triangle  BDE\ 
calling  BD  =  hv  the  total 
pressure  represented  by  the 
triangle  BDE  is  expressed 
by  P  =  l  hfy.  Thus,  in  the 
case  of  water  pressure  as  in 


B 
P 


/7  ^  the  problem   of    earth    pres- 

FlG>  65t  sure,   the    total    pressure    is 

proportional  to  the  square  of  the  height  of  the  wall.  In 
the  triangle  ABC,  the  ordinates  measured  parallel  to  A  C 
represent  the  intensities  of  pressure  at  the  corresponding 
depths;  thus  the  length  of  the  line  AC  represents  the  in- 
tensity of  pressure  at  the  depth  BA,  while  the  length  of  the 
line  DE  represents  the  pressure  per  square  foot  at  a  depth 
BD.  The  total  pressure  on  a  small  element  of  height  near  the 


DAMS 


109 


point  D  is  therefore  measured  by  the  area  of  the  strip  between 
the  ordinates  drawn  at  the  upper  and  lower  edges  of  the  ele- 
ment, which  is  equal  to  the  mean  ordinate  DE  multiplied  by 
the  height  of  the  element.  The  triangle  ABO  is  the  sum  of 
all  the  pressure  areas  for  all  the  successive  elements  of  height 
composing  the  wall  AB,  or  is  equal  to  the  total  pressure.  It 
follows  that  the  resultant  pressure  acts  through  the  center  of 
gravity  of  triangle  ABC,  which  is  located  one  third  of  AB 
above  the  base  A.  The  location  of  this  resultant  is  called 
the  center  of  pressure. 

The  location  of  the  center  of  pressure  can  also  be  deter- 


mined analytically,  as  follows: 

Represent  any  portion  of  the  back  of 
the  dam  by  the  area  ABCD  (Fig.  66). 
The  coordinates  of  points  of  this  area  will 
be  denoted  by  x  and  y,  measured  from  0, 
in  the  water  surface,  y,  representing  the 
horizontal  distance  from  the  center  line  of 
the  area,  and  x  the  depth  from  the  water 
surface  down  to  the  point  in  question. 
Consider  a  narrow  vertical  strip  of  width 
y.  The  center  of  pressure  of  this  strip  is 
at  a  depth  X  below  the  surface,  given  by 


B  yo c 


D 


FIG.  66. 

the  following  formula,  for  a  total  height  of  the  strip  equal 
to  h : 


X  = 


Ch 

\    yxdx 


Since  y  is  constant,  we  may  integrate  for  the  rectangular 
area  of  the  strip,  thus, 


— , 
o 


110    EARTH   SLOPES,   RETAINING    WALLS,  AND  DAMS 
and 


so  that  we  have 

f 


or  the  center  of  pressure  is  at  f  the  total  height  of  the  strip 
from  the  surface  of  water,  or  at  \  the  height  from  the  base. 
The  whole  area  ABCD  is  composed  of  a  large  number  of 
such  strips  side  by  side,  as  the  center  of  pressure  for  each 
is  at  the  same  height,  the  resultant  center  of  pressure  is 
also  at  J  h  above  the  base. 

Dams  fail  from  precisely  the  same  causes  as  retaining 
walls.  Like  a  retaining  wall,  a  dam  may  fail  in  any  one 
of  three  different  ways,  viz.,  by  rotation  or  overturning,  by 
sliding,  and  by  crushing.  To  insure  the  stability  of  a  dam, 
three  conditions  must  therefore  be  satisfied  : 

1.  The  resultant  of  water  pressure  and  weight  must  fall 
within  the  middle  third  of  the  base.     This  will  give  a  factor 
of  safety  of  2  against  overturning,  or  rotation  around  the 
exterior  point  of  the  base. 

2.  The   angle   made  by  the    resultant  with  the  perpen- 
dicular to  the  base  must  be  smaller  than  the  angle  of  friction 
between  the  foundation  and  the  dam.     This  condition  will 
insure  stability  against  sliding. 

3.  The  maximum  intensity  of   pressure    at   the  base  of 
the  dam   must  be  less  than  the  crushing  strength  of  the 
material.     This  condition  will  insure  safety  against  crush- 
ing.    A  suitable  factor  of  safety  must  be  employed  here. 


DAMS  111 

The  same  conditions  must  be  satisfied  at  every  horizontal 
section  of  the  dam,  since  each  section  BDE  (Fig.  65)  may 
be  considered  as  an  independent  dam  resting  on  the  part 
ADEC  as  a  foundation. 

THEORETICAL  PROFILE  FOR  DAMS 

52.  In  studying  dams  we  will  always  deal  with  a  slice 
one  foot  in  length.     Then  the  area  of  the  cross-section  is 
numerically  equal  to  the  volume  of  the  portion  considered, 
and  the  area  of  the  triangle  of  pressure  multiplied  by  the 
weight  of  a  cubic  foot  of  water  is  equal  to  the  total  pressure. 

53.  Triangular  Profile.  —  The  most  economical  profile  for 
a  dam,  theoretically,  is  a  triangular  one,  with  downstream 
slope  as  given  by  the  formula,  demonstrated  below, 


where  <ya  and  7^  represent  the  weights  of  the  unit  of  volume 
of  water  and  masonry  respectively,  t  the  thickness 
of  the  dam  at    the  base  and  h  the  total  height  of 
the  structure  (Fig.  67). 

The   pressure   P  of  water  against  the  dam 
when  the  water  surface  is  level  with  the  crown 
is  given  by  the  formula, 


Hydraulics  teaches  that  the  pressure 

of  water  on  a  submerged  surface  of 

any  shape  whatever  equals  the  area 

of   the  surface  multiplied  by  the 

vertical  distance  from  its  center 

of  gravity  to  the  surface  of   the 

water,   multiplied    by   the  weight  FIG.  67. 


112     EARTH  SLOPES,   RETAINING  WALLS,   AND  DAMS 

of  a  cubic  foot  of  water.  In  the  present  case  the  water  side 
of  the  dam  being  vertical,  and  the  area  of  wetted  surface 
being  a  rectangle  whose  width  is  one  foot  and  whose  height 
is  equal  to  the  depth  of  water,  the  center  of  gravity  of  the 

wetted  area  will  be  at  the  middle  point  of  the  height,  or  at  - 

below  the  water  surface,  while  the  area  is  1  x  h  =  h  ;  the 
pressure  P  will  be,  therefore, 


The  weight  of  the  dam  (Fig.  67)  will  be, 

W=lthym. 
Dividing  the  pressure  by  the  weight, 


%nt 

The  outer  edge  of  the  middle  third  of  the  base  will  be  at 
1  1  from  W,  since  the  vertical  line  through  the  center  of 
gravity  of  a  right  triangle  will  meet  the  base  at  ^  the  width 
from  the  right  angle.  Then  taking  moments  about  the  outer 
edge  of  the  middle  third  of  the  base,  the  resultant  moment  must 
be  equal  to  zero  if  the  resultant  pressure  passes  through  the 
edge  of  the  middle  third  as  is  demanded  by  the  condition  of 
stability  against  overturning  ;  that  is, 

P  x  iA-TTx^  =  0, 
or 


from  which  we  find  the  relation  necessary  for  stability  to  be, 

P__l±_t_ 
W~h~  h 


DAMS 


113 


But  we  found  above  that 


P 

W 


7m* 


jf 

Equating  the  two  expressions  for        gives 


or 


t 

h 

A2 
h, 


7« 

7m' 


and  consequently 


7m 


Therefore  -j,  which  is  equal  to  the  slope  of  the  down- 
stream face  of  the  dam,  must  be  made  equal  to  the  square 
root  of  the  specific  weights  of  water  and  masonry.  Conse- 
quently the  thickness  t  of  the  base  of  the  triangular  dam 
will  be  given  by  /*/•'/? 


54.    Trapezoidal  Profile.  —  In  actual  prac- 
tice it  is  impossible  to  use  the  triangular 
dam  profile,  since  it  is  necessary  to  give 
the  dam  a  certain  thickness  at  the  top. 
Therefore  the  simplest  profile  is  the 
trapezoidal  one,  as  exhibited  by  Fig. 
68.     Here  ABCD  is  the  cross-sec- 
tion of  the  dam,  t1  being  the  thick- 
ness of  the  structure  at  the  top. 
Using  the  same  notation  as  in  the  ^7 
preceding    case,  the   pressure   of 


FIG.  68. 


114     EARTH  SLOPES,    RETAINING  WALLS,  AND   DAMS 

the  water  is 

P  =  \  V& 
and  the  weight  of  the  dam  is 

Wi***l(t+t')fym, 

The  weight  Wean  be  considered  as  the  sum  of  the  weights 
of  the  rectangular  prism  ABOE  and  the  triangular  prism 
DEO.  Call  the  weights  of  these  two  prisms  W1  and  W%. 
Then  we  will  have 


and  Wi=$Vm(t  +  tr)h. 

Let  F  be  the  outer  edge  of  the  middle  third  of  the  base 
AD.     The  distance  of  Wl  from  F  will  be 


while  the  distance  of  W2  from  F  will  be 

..  pw  =  2 1  _  tf  _  t_-^_ _  2t-3j^-t±t' 

_t-2tf 
^2-    — 

For  equilibrium  about  the  point  F  we  must  have  the  alge- 
braic sum  of  the  moments  equal  to  zero,  or 

P  =  i  h  -  Wlw1  -  W2w2  =  0. 

Substituting  in  this  equation  the  various  values  found  above, 
we  obtain 


Simplifying, 

-  3  O  -  i  7.(«  -  f)*(*  -2O-  0- 


DAMS  115 

Now  write  at  in  place  of  £',  so  as  to  represent  t'  in  terms  of 
£,  a  being  a  coefficient.     Then 

1  7aA3  -  1  ««A(4  1  -  3  «0  -  J  7w(*  -  «OA(«  -  2  O  =  0. 
This  reduces  as  follows  : 

-  3  «*)  -  7,»0  -  O(*  -  2  «0  =  °» 

22    =  Q, 


or 


which  is  equal  to  the  thickness  of  base  required  for  a  pro- 
file, as  found  above,  divided  by  the  quantity  Vl  -f  a  —  a2. 
For  both  the  values  a  =  0  and  a  =  1  the  quantity  Vl  4-  a  —  a2 
is  equal  to  1.  This  means  that  a  triangular  and  a  rectangu- 
lar profile  require  the  same  base  width.  In  passing  from 
the  one  to  the  other,  the  quantity  referred  to  increases  to  a 
certain  limit  from  which  it  decreases  again.  Its  greatest 
value  corresponds  to  a  =0.5.  This  means  that  the  thick- 
ness t  of  the  base  of  the  dam  decreases  with  increase  of  a 
from  0  to  0.5,  while  it  increases  when  the  value  of  a  lies 
between  0.5  and  1.  In  other  words,  it  will  be  most  eco- 
nomical to  give  to  the  top  of  the  dam  a  thickness  less  than 
half  the  width  of  the  base  ;  greater  thickness  of  top  produces 
a  waste  of  masonry. 

55.  Pentagonal  Profile.  —  In  the  trapezoidal  profile  there 
is  an  excess  of  stability  at  the  section  near  the  top  of  the 
dam,  which  means  that  a  large  quantitj^  of  masonry  is  used 
that  might  be  saved  by  a  better  shape  of  cross-section. 


116     EARTH   SLOPES,   RETAINING  WALLS,   AND  DAMS 


Since   the   best   theoretical   profile   is   the    triangular   one, 
which  is  objectionable  only  because  it  gives  zero  thickness 

at  the  top,  we  may  retain  approxi- 
mate theoretical  perfection  and 
yet  correct  this  defect  by  using 
a  cross-section  whose  lower  part 
is  similar  to  the  corresponding 
part  of  the  triangular  section 
while  the  upper  part  is  a  wall 
of  constant  thickness.  The  result 
of  this  modification  is  the  pentag- 
onal profile,  Fig.  69. 

ABODE,  the  profile  of  the  dam, 
can    be    considered    as   composed 
of  the  triangle  ABE  of  altitude  h 
and  base  t,  and  the  triangle  BCD 
FlG-  69-  of  altitude  hr  and   base  t'  .     The 

pressure  of  water  will  be  as  before, 


% 


the  weight  of  the  dam  will  be  equal  to  the  weight  of  the 
two  triangular  prisms,  W—  W1 

since  Wl  =  J  7 

and  TTa  =  j7 

we  will  have 


The  right  triangles  AEB  and  BCD  are  similar,  having 
their  angles  equal  and  their  sides  parallel  ;  consequently 
their  sides  will  be  proportional, 


DAMS  117 

or 


Expressing  the  dimensions  of   the  upper  section   BCD   in 
terms  of  the  width  of  base  by  making  t'  =  at,  we  obtain, 


or 


For  stability  against  overturning  it  is  necessary  that  the 
algebraic  sum  of  the  moments  about  the  outer  edge  of  the 
middle  third  of  the  base  be  zero.  That  is,  using  the  dis- 
tances w1  and  w2  in  the  same  sense  as  before, 

P  x  J  h  -  W1wl  -  TJ>2  =  0. 
Now 

wl  =  %t  and  w2  =  FA  -  NA  =  f  *  -  f  at  =  f  £(1  -  a), 
so  that  the  equation  of  moments  becomes, 

\  7a/*2  X  \  h  -  1  ymth  X  J  *  -  1  yma*th  X  f  *(1  -  a)  =  0, 
or,  reducing, 

t  7^3  -  |  7^  -  1  7»«2(1  -  «)«  =  0, 
or 

7a^2  -  7»A1  +  2  «2(1  _  «))  =  0, 
from  which 


or 


A, 


118     EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 

which  is  the  required  thickness  of  base  for  the  pentagonal 
profile. 

Dividing  both  terms  by  h, 


t_     : 

h~  Vl  +  2«2-2a3' 

which  is  the  slope  EB.  It  is  equal  to  the  slope  of  the  pure 
triangular  profile,  "\— ,  divided  by  the  quantity  Vl  +  2  a2  —  2  a3. 

This  quantity  is  equal  to  1  when  «  =  0  and  when  a  =  1,  i.e. 
for  the  pure  triangular  and  the  pure  rectangular  profile. 
By  differentiation,  Vl  +  2  a2  —  2  a3  with  respect  to  a,  we 
may  find  what  value  of  a  gives  the  greatest  value  to  the 
denominator,  and  consequently  the  smallest  width  of  base. 
It  is  found  to  be  0.666,  that  is,  the  smallest  bottom  width  t 
is  obtained  when  £',  the  top  width,  is  made  two  thirds  of  the 
bottom  width. 

56.  In  the  following  tables  the  various  dimensions  and 
values  for  the  minimum  trapezoidal  and  pentagonal  profiles 
are  given  in  terms  of  the  height.  In  working  out  these 
tables  the  ratio  of  upper  base  to  lower  base  was  taken  at 
successive  tenths  from  0  to  1.  The  weights  of  the  materials 
were  assumed  to  be 

water,        ja=  62.5  Ibs.  per  cubic  foot. 

masonry,  <ym  =  150  Ibs.  per  cubic  foot. 


DAMS 


119 


II 

H  H 


1 


M 
d 

jo  XKaioujaoQ 


»O  Ci  T^ 
^H  CO  CO 
CO  CO  CO 


O5  CO  O  <M 

m  £J  fc:  £» 

T  ^P  CO  CO 


»0  "*  "*  t-  ^  "*  ^  t-  10  ^  * 

oooooJT-^iOGOC'it^co'^Hcd 


lO  ^O  ^-O  ^O  K71  *O  ^O  ^O  ^O  >O  IO 
OJ  OJ  (Ol  (01  Ol  (>1  Ol  CJ  CJ  CJ  Ol 

CO  CO  CO  CO  CO  CO  CO  CO  CO  CO  CO 


§3 


01  o  co  T^  "^ 

Oi  CO  Ol  Ol  iO 


^O1'*O5CO<^'— ICOOO-<* 
O  O  rH  i— I  Ol  OJ  CO  CO  ^  O  CO 

ooooooooooo 


'asrg 


•*    CO    ^    CO 
(M    i-l    <M    CO 


ooooooooooo 


»=-  'oixvg 


O'-jOlCO1<^^OCC5t>.OOOSO 
OOOOOOOOOOO 


»— (OlCO<^l»OCDi>.OOC5O«— i 


120      EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 


£3   ^ 

ce 

tt 

« 
1 

2  £  " 
II, 

el 

1 

II"  • 
££ 

-   j§ 

1 

A      .soiiOlH.! 

•I 

lO    r^    ^""^    OS 
CO    CO    CO    CO 

OS     rH     CO     O     Ol     T* 

co  10  ^o  ^o  *o  ^^ 

CO    CO 

'iiioia^ 

CT        (M        «        C^ 

la     °*        °l»     °1»     °i»     "* 

CO     rH 
CP     GO 

GO    l~*-    GO    OS 

O    CO    b-    O    Ol    O 
O    O    lO    CO    CO    t— 

rH     CO 

GO    OS 

*4£-y 

Ol    01    Ol    01 

(M        0»        IM        (M        <X        C-J^ 

lO    »O    >O    >O    >O    >O 
Ol    Ol    Ol    Ol    0^1    Ol 

®i     e» 
Ol    Ol 

'aaassaaj  aaxv^vi 

CO    CO    CO    CO 

CO    CO    CO    CO    CO    CO 

CO    CO 

7, 

IM        O»        <M        »N 

IN      <M      ej^    w      c«      e^ 

Ss  3s 

¥2(z    +  DJ    J. 

Ol    ^H    Ol    CO 

CO    CO    CO    CO 

CO    GO    CO    -^    ^    GO 
CO    >O    GO    O    i—  (    CO 
CO    CO    CO    ^    -^    ^ 

S    CO 

jo9M^ 

O     O     r-l     i—  ( 

Ol    >O    CO    OS    Ol    CO 
CO    GO    CO    CO    OS    »O 
O-l    Ol    CO    CO    CO    ^?t* 

to    co 

2  'asvg 

»O    CO    O    !>. 

Tt<    CO    Ol    O 
O    CO    CO    CO 

<O    CO    ^^    Ol    CO    Ol 

t^    iO 

OS    ^^ 
IO    CO 

i 

O    »—  1    <OI    CO 

^H      iQ     C^5      C^     ^>^     QQ 

os  o 

0-000 

o  o  o  o  o  o 

O     rH 

-- 

rH    <M    CO    ^ 

O    O    t--    GO    OJ    O 

rH     rH 

DAMS  121 

PRACTICAL  CKOSS-SECTIONS 

57.  The  theoretical  dam  profiles  examined  in  the  preceding 
chapter  are  very  seldom  used  for  practical  purposes.     But 
the  trapezoidal  and  pentagonal  profiles  are  valuable  guides  to 
the  engineer  in  designing  suitable  cross-sections  for  dams. 
The  trapezoidal  profile  approximates  the  preferred  outline 
for  submerged  dams,  while  the  pentagonal  profile  is  the  basal 
form  of  all  the  high  dams  erected  in  recent  years. 

58.  Submerged  dams.      Ogee  Profile.  —  According  to  Pro- 
fessor Merriman,  the  pressure  of  water  against  a  submerged 
dam  is  given  by  the  formula 

P=£A(A  +  <f), 

where  d  is  the  height  of  water  above  the  dam.  The  point  of 
application  of  the  resultant  pressure  on  the  upstream  side  of 
the  dam  is  at  a  height  above  the  base  given  by  the  formula, 

h  +  3d     1    , 

•  ;•;  .      ^=ATTdx3<*-  1 

The  most  suitable  profile  for  a  submerged  dam  is  the  one 
having  an  ogee  curve  on  the  downstream  side  which  guides 
the  water  flowing  over  the  crest.  Such  a  profile  is  produced 
by  changing  the  downstream  face  to  the  form  of  a  reverse 
curve  tangent  to  the  face  at  mid-height,  the  upper  curve 
giving  a  rounded  top  to  the  dam  while  the  lower  one  turns 
forward  horizontally  downstream. 

In  designing  a  submerged  dam  where  the  depth  of  water 
over  the  crest  will  not  be  greater  than  10  ft.,  the  following 
practical  rule  may  be  used  with  advantage.  It  gives  a  profile 
in  which  the  line  of  resistance  is  clearly  shown  in  Fig.  71, 
and  it  is  very  close  to  the  median  line  of  the  structure. 

Determine,  first,  the  thickness  of  base  for  a  trapezoidal 


122      EARTH   SLOPES,   RETAINING   WALLS,   AND  DAMS 


profile  according  to  the  method  given  previously,  taking  as 
depth  of  water  the  height  of  dam  plus  the  depth  of  water 
over  the  crest,  or  H  =  h  -f  d,  and  making  the  top  width  0.3 
of  the  width  of  the  base;  in  other  words,  making  a  =  0.3. 
The  trapezoidal  profile  AS  DO  (Fig.  70)  will  thus  be  obtained. 

I  Bisect  the  angle  A  CD  by  the 
line  CM.  Then  from  the  mid- 
dle point  of  AC  drop  the  verti- 
cal N  meeting  the  bisector  at 
M.  With  M  as  center  and 
MN  as  radius  describe  the  arc 
NP.  Similarly  bisect  the  angle 
EDF  by  the  line  DO.  From 
i  a  point  R  on  the  back  of  the 
dam,  at  a  height  equal  to  4d 
above  the  base,  draw  a  horizon- 
tal line  RO  meeting  the  bisec- 
tor at  0.  With  0  as  center 
and  OE  as  radius  describe  the 
arc  EQF.  The  result  is  the  profile  ANPFQEB,  which  is  a 
suitable  cross-section  for  submerged  dams  when  the  height 
of  water  above  the  dam  or  d  is  not  more  than  10  ft. 

The  resistance  of  an  ogee  profile  designed  according  to 
the  preceding  rule  was  determined  graphically  by  Mr. 
William  J.  S.  Deevy,  and  the  result  of  his  work  is  given  in 
the  following  table,  referring  to  Fig.  71.  The  data  assumed 
were  as  follows: 

h  =  height  of  dam  =  80  ft. 

d  =s  depth  of  water  over  the  crest  =  5  ft. 

7a  =  weight  of  water  per  cubic  foot  =  62^  Ibs. 

7TO.  =  weight  of  masonry  per  cubic  foot  =  150  Ibs. 


FIG.  70. 


DAMS 


123 


FIG.  71. 


6     H 

g 

W     H 
-     k 

1 

a  sr 

•<   § 

•    » 
-  S 

*  8 

«§s 

rjTj 

£  I 

0 

s  g  . 

§ 

H 

o  g 

0    g 

£    H 

o  >  § 

hi 

H     *     ~ 

^ 

CO 

T  5 

O          BS 

^     2    g^ 

o   g 

S 

&*» 

h 

h     W 
0   ifi 

*l 

o 

SB." 

J*g 

«  1 

a 

«  S  | 

.    °  % 

=  « 

a  a. 

»  2 

£>   E^    ._ 
JB    S5    fc 

z  fe 

M 
U     O 

1 

s:i 

d     H 

o   £ 

0     ° 

S   S  tj 

i  «  8 

ISM 

H    %    g 

fc     H 

M 

fe  ^ 

«<     1 

^  *  >3 

K   "  £ 

Q     H    fe 

p  §0^ 

Q    H    fafe 

1 

21.5 

6250 

173.33 

26000 

4.16 

10.5 

11.5 

9 

2 

26 

18750 

410.83 

61625 

11 

13 

13 

10.5 

3 

31 

37500 

695.83 

104375 

11.25 

15.5 

15 

12 

4 

36 

62500 

1030.83 

154625 

14.66 

18 

16.7 

13.5 

5 

41 

93750 

1415.83 

212375 

18.05 

20.5 

17.5 

14.5 

6 

46 

131250 

1850.83 

277625 

21.43 

23 

19.5 

16 

7 

50.5 

175000 

2333.33 

350000 

24.8 

25.75 

20.5 

17.5 

8  base 

67.5 

224000 

2923.33 

438500 

28.1 

33.75 

35.5 

19.5 

59.  High  Dams.  —  Many  authors  have  suggested  practical 
rules  for  the  designing  of  high  dams.  All  these  profiles, 
however,  closely  resemble  the  pentagonal  profile,  with  the 
difference  that  the  length  of  the  horizontal  joints  is  increased 
either  by  giving  a  batter  to  the  upstream  side  of  the  dam 
or  by  making  the  face  curvilinear,  while  on  the  downstream 


124      EARTH   SLOPES,   RETAINING  WALLS,   AND  DAMS 


face  the  vertical  and  inclined  portions  are  connected  by  a 
curve  of  small  radius.  In  such  profiles  the  line  of  resistance, 
both  with  full  and  with  empty  reservoir,  falls  within  the 
middle  third  of  the  joint  at  all  joints. 

60.    Crugnola's  Section.  —  Mr.  Giacomo  Crugnola,  in  his 
work,  "  Sui  muri  di  sostegno  delle  terre  e  sulle  traverse  dei 

serbatoi  d'acqua," 
gives  the  following 
data  for  designing 
masonry  dams,  re- 
sulting in  the  pro- 
file shown  by  Fig. 
72.  In  this  table 
the  various  dimen- 
sions by  which  the 
profile  is  deter- 
FlG*  72*  mined  are  given  in 

meters,  and  the  volumes  of  masonry  are  given  in  cubic 
meters. 


UPSTREAM  SIDE 

DOWNSTREAM  SIDE 

A 

a 

Sc 

P,n 

Km 

F 

Pv 

Rv 

t 

r 

s 

ft 

VOLUME 

5 

0.5 

1.7 

3.5 

16 

0.07 

1.7 

4 

2.75 

5 

2.52 

10.500 

10 

0.9 

2 

6.0 

24 

0.34 

1.8 

6 

4.75 

10 

6.08 

35.055 

15 

1.3 

2.3 

7.0 

32 

1.02 

1.9 

8 

6.00 

15 

9.81 

76.637 

20 

1.5 

2.5 

8.0 

40 

1.84 

2.0 

10 

7.25 

20 

13.70 

133.010 

25 

2 

3.00 

9.0 

48 

2.75 

2.1 

12 

8.50 

25 

17.99 

217.700 

30 

2.4 

3.5 

10.0 

56 

3.69 

2.2 

14 

9.50 

30 

21.75 

314.557 

35 

2.8 

4.0 

11.0 

64 

4.47 

2.3 

16 

11.50 

35 

27.90 

455.804 

4Q 

3.00 

4.25 

12.0 

72 

5.67 

2.4 

18 

13.0 

35 

34.04 

1 

610.442 

45 

3.25 

4.50 

13.5 

80 

6.46 

2.5 

20 

14.5 

35 

38.88 

1 

781.423 

50 

3.50 

4.75 

15.0 

88 

7.26 

2.6 

22 

16.0 

35 

46.92 

1 

996.108 

DAMS 


125 


61.  Krantz's  Section.  —  The 
French  engineer  Krantz,  in  his 
book,  "Etude  sur  les  murs 
de  reservoirs,"  has  suggested 
another  practical  profile  for 
designing  dams  of  various 
heights.  Krantz's  profile  is 
shown  in  Fig.  78.  All  the 
elements  for  designing  the 
Krantz  profile  for  any  height 
are  given  in  the  following  table  in  meters : 


A 

it 

Ji 

R 

VOLUME 

*5 

Jli 

5 

2.0 

4 

13 

13 

14.280 

10 

2.5 

7 

26 

21.35 

42.190 

15 

3.0 

10.5 

39 

27.25 

85.960 

20 

3.5 

14.5 

52 

32.07 

147.660 

25 

f  "'""* 

4.;o 

19 

65 

36.25 

229.120 

30 

4.5 

24 

78 

40.08 

331.890 

35 

5.0 

29.5 

91 

43.75 

457.290 

40 

5 

5.0 

29.5 

1 

1 

5 

3.33 

91 

43.75 

635.620 

45 

10 

5.0 

29.5 

1 

1 

10 

6.67 

91 

43.75 

855.640 

50 

15 

5.0 

29.5 

1 

1 

15 

10 

91 

43.75 

1117.290 

Both  Crugnola  and  Krantz  determined  their  profiles  by  a 
long  series  of  trials  designed  without  the  use  of  any  formula 
or  definite  rule. 

62.  Author's  Section. — A  practical  profile  for  a  dam  of 
any  height  may  be  easily  obtained  in  the  following  manner, 
based  upon  the  theoretical  profile.  The  method  can  be  ap- 
plied to  any  case. 

Calculate  the  thickness  of  base  for  a  pentagonal  profile 
(Fig.  69)  by  the  methods  previously  given,  assuming  for  a 
the  value  of  0.2,  so  that  tr  =  0.2 1.  From  the  reentrant  angle 


126      EARTH   SLOPES,   RETAINING   WALLS,   AND   DAMS 


of  the  downstream  face,  where  the  vertical  and  inclined  por- 

tions  of  the  face 
meet,  draw  a  hori- 
zontal line,  and  with 
center  on  this  line, 
upstream  from  the 
dam,  and  a  radius 
equal  to  three  times 
the  height  of  the 
dam,  describe  an  area 
of  circle  tangent  to 
the  upstream  face ; 
this  curve  will  form 
FIG.  74.  the  back  of  the  dam 

in  place  of  the  vertical  face  of  the  theoretical  profile. 
Then  join  the  vertical  and  inclined  portions  of  the  down- 
stream face  by  a  curve  of  radius  equal  to  the 
top  width  of  the  dam.  There  results  a  pro- 
file similar  to  the  one  shown  in  Fig.  74. 

The  lines  of  resistance  of  a  dam  100  ft. 
high   designed   according  to   the  method 
indicated  above  are  given  in  Fig.   75, 
both  for  reservoir  full  and  for  reser-  /      j/L 

voir   empty.     In   the   calculation   of  ~ 

these   lines,  the   weight   of   water  — 

was  assumed  to  be  625  Ibs.,  and 
the  weight  of  masonry  150  Ibs.        / 
per  cubic  foot.     The    dimen- 
sions   and    pressures    for  the 
various    joints,    as    given    in 
the  following  table,  were  cal-  FJG-  75. 

culated   by    Mr.   John  M.    Fitzgerald: 


DAMS 


127 


s 

s 

2 

j 

*§s 

a     ** 

^n 

b 

GO 

£ 

o 

«  H  ^  «" 

§  o  as 

ill 

I 

• 

1-3 

^  °  *  « 

b     O 

*l 

h 

o 

H  |  0 

H  oj  eJ 

IB 

C£ 

H 

2                Q      W 

B  l<  M  M 

fe  P5 

| 

if 

M 

sis*  . 

ill" 

sl^sf 

00 

^3 

F5 

3<E 

pifi«S*£ 

CCSO& 

(55£fc 

1 

3115 

22500 

13.5 

6.0 

0.5 

7.0 

2 

12500 

45250 

16.5 

7.0 

2.5 

7.5 

3 

28125 

72500 

19.5 

7.0 

4.5 

8.0 

4 

50000 

107500 

27 

11.5 

6.0 

9.5 

5 

78125 

155000 

34 

14.5 

7.5 

12 

6 

112500 

207.500 

42 

16.5 

10.5 

15.0 

7 

153125 

280.000 

49.5 

19.0 

12.5 

18.0 

8 

200000 

350000 

56.5 

21.0 

15.5 

20.0 

9 

253.125 

420000 

64.5 

23.5 

17.0 

24.0 

10 

312.500 

556250 

73.5 

25.5 

18.0 

34.0 

Comparing  the  figures  in  the  last  three  columns,  it  will  be 
seen  that  the  lines  of  resistance,  both  reservoir  full  and 
reservoir  empty,  falls  well  within  the  middle  third  of  the 
corresponding  joint. 

In  very  high  dams  the  value  of  a  for  this  profile  (a  =  t'-r- 1, 
or  top  width  divided  by  width  of  base)  may  even  be  decreased 
to  0.15  or  0.1;  if  this  is  not  done,  the  dam  will  be  very  thick 
at  the  top,  with  corresponding  waste  of  masonry. 

Dams  built  for  reservoir  purposes  must  be  higher  than  the 
level  of  water  in  the  pool  formed  by  the  dam.  Tne  portion 
of  the  structure  which  rises  above  the  water  is  called  the 
crest  of  the  dam.  It  is  noteworthy  that  both  Krantz  and 
Crugnola,  as  well  as  all  others  who  have  suggested  dam 
profiles,  have  given  various  dimensions  for  the  height  of 
crest  of  dams  of  various  heights.  As  a  general  rule,  it 
is  well  to  make  the  crest  at  least  ^  the  total  height 
of  the  dam.  In  the  profile  suggested  by  the  author,  the 
crest  was  not  taken  into  account,  but  following  the  general 


128      EARTH  SLOPES,   RETAINING   WALLS,   AND  DAMS 

rule  it  would  be  very  easy  to  add  to  the  profile  of  Fig.  75 
a  prism  of  masonry  10  ft.  high.  Since  no  water  pressure 
is  added  hereby,  the  added  prism  of  masonry  tends  to  in- 
crease the  stability  of  the  structure,  since  it  increases  the 
weight  upstream  of  the  center  of  gravity. 

In  regard  to  the  manner  of  determining  graphically  the 
line  of  resistance  in  a  dam  of  a  given  profile,  the  following 
method  may  be  used:  divide  the  profile  into  a  number  of 
sections  by  drawing  horizontal  lines  5  ft.  or  10  ft.  apart. 
These  lines  will  divide  the  profile  into  so  many  dams  with 
parallel  bases,  in  each  case  considering  the  entire  area  above 
the  horizontal  line  or  joint  in  question.  Find  the  center  of 
gravity  of  each  of  these  small  dams  and  draw  vertical  lines 
through  the  various  centers  of  gravity.  Mark  the  point 
where  each  vertical  cuts  its  corresponding  base,  and  draw  a 
continuous  line  through  the  points  thus  obtained.  This  is 
the  line  of  resistance  for  empty  reservoir,  or  when  the  dam 
has  no  pressure  to  resist.  Next  draw  a  force  polygon,  laying 
off  vertically  in  succession  the  weight  of  the  successive  slices 
of  the  dam.  From  the  upper  end  of  the  first- weight  line  in 
the  force  polygon  draw  a  horizontal  line  and  upon  it  lay  off, 
to  the  same  scale  as  used  for  the  weights,  the  successive 
water  pressures  on  the  various  slices.  Connect  the  ends  of 
the  pressure  and  weight  lines  corresponding  to  each  other. 
The  connecting  lines  represent  to  scale  the  resultants  of 
weight  and  pressure  for  the  several  sections  of  the  dam,  in 
each  case  the  entire  section  above  the  joint  considered. 
Then  return  to  the  profile,  and  at  J  of  the  height  of  each 
section  or  partial  dam  (point  of  application  of  the  pressure) 
draw  from  the  vertical  which  was  dropped  from  the  center 
of  gravity  a  line  parallel  to  the  corresponding  resultant. 
Mark  the  points  where  the  various  resultants  intersect  the 


DAMS 


129 


corresponding  points,  and  connect  these  points  by  a  con- 
tinuous line.  This  is  the  line  of  resistance  for  full  reservoir, 
i.e.  when  the  dam  is  resisting  the  maximum  water  pressure. 
The  dash-and-dot  lines  in  P^ig.  75  are  the  two  lines  of  resist- 
ance, the  right-hand  one  being  for  reservoir  empty  and  the 
left-hand  one  for  reservoir  full.  The  gravity  lines  and  the 
lines  of  resultant  pressure  for  each  dam  of  partial  height  are 
also  shown  in  the  figure,  showing  clearly  how  the  lines  of 
resistance  are  obtained. 


9?  THE 

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appendix,  tables,  tests,  and  formulse  for  the  use  of  Electrical  Engineers. 
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REINHARDT,  CHAS.  W.     Lettering  for  Draughtsmen,  Engineers  and 

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STALEY,  CADY,  and  PIERSON,  GEO.  S.     The  Separate  System  of 

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THURSO,  JOHN  W.     Modem  Turbine  Practice  and  Water-Power 

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TOWNSEND,  F.  Short  Course  in  Alternating  Current  Testing.  8vo, 
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URQUHART,  J.  W.  Dynamo  Construction.  A  practical  handbook 
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WEISBACH,  JULIUS.  A  Manual  of  Theoretical  Mechanics.  Ninth 
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WILSON,  GEO.  Inorganic  Chemistry,  with  New  Notation.  Revised 
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WRIGHT,  Prof.  T.  W.  Elements  of  Mechanics,  including  Kinematics, 
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Return  to  desk  from  which  borrowed. 
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NOV19   1947 


REC'D  ED 

OCT241932 


LD  21-100m-9,'47(A5702sl6)476 


'  UXr\ 


